a) the cost of the house is 96000 and the down paidment is the 20% so we can use a rule of 3 to solve it so:
[tex]\begin{gathered} 96000\to100 \\ x\to20 \end{gathered}[/tex]so the equation will be:
[tex]\begin{gathered} x=\frac{96000\cdot20}{100} \\ x=19200 \end{gathered}[/tex]b) her income is $4000 so the 28% will be:
[tex]\begin{gathered} 4000\to100 \\ x\to28 \end{gathered}[/tex]so the equation will be:
[tex]\begin{gathered} x=\frac{4000\cdot28}{100} \\ x=1120 \end{gathered}[/tex]c) the equation that models a loan is:
[tex]C=\frac{P\cdot(0.095\cdot(1+0.095)^n)}{(1+0.095)^{25}-1}[/tex]So we replace the princeiple and find monthly paidment.
[tex]\begin{gathered} C=\frac{76800\cdot(0.095\cdot(1.095)^{25}}{8.67} \\ C=\frac{76800\cdot0.92}{8.67} \\ C=\frac{70540.30}{8.67} \\ C=8136.14 \\ C\approx8136 \end{gathered}[/tex]d)The total monthly paidment will be:
[tex]\begin{gathered} T=8136+\frac{1300}{12}+\frac{336}{12} \\ T=8136+108.33+28 \\ T=8272.33 \\ T\approx8272 \end{gathered}[/tex]SOo the answer is NO she can't afort to buy this house
[tex](x-1)(x^{2}+2)[/tex]
Answer:
x³-x²+2x-2
That's the answer
Let f(x) = x² + 11x + 25 Find a so that f(a) = 1
A=-3
A=-8
Explanation
Step 1
[tex]f(x)=x^2+11x+25[/tex]there is a number A so f(A) =1, then
[tex]\begin{gathered} f(A)=A^2+11A+25 \\ f(A)=1 \\ \text{then} \\ A^2+11A+25=1 \\ A^2+11A+24=0\text{ equation(1)} \end{gathered}[/tex]Step 2
solve using the quadratic equation
[tex]\begin{gathered} \text{for } \\ ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]a)let
a=1
b=11
c=24
the variable is A,
b) replace
[tex]\begin{gathered} A=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ A=\frac{-11\pm\sqrt[]{121^{}-4\cdot1\cdot24}}{2\cdot1} \\ A=\frac{-11\pm\sqrt[]{121^{}-96}}{2} \\ A=\frac{-11\pm\sqrt[]{25}}{2} \\ A_1=\frac{-11+\sqrt[]{25}}{2}=\frac{-11+5}{2}=\frac{-6}{2}=-3 \\ A_1=-3 \\ A_2=\frac{-11-\sqrt[]{25}}{2}=\frac{-11-5}{2}=\frac{-16}{2}=-8 \\ A_2=-8 \end{gathered}[/tex]I hope this helps you
A gardener builds a rectangular fence around a garden using at most 56 feet of fencing. The length of the fence is four feet longer than the widthWhich inequality represents the perimeter of the fence, and what is the largest measure possible for the length?
We know that
• The gardener used at most 56 feet of fencing.
,• The length of the fence is four feet longer than the width.
Remember that the perimeter of a rectangle is defined by
[tex]P=2(w+l)[/tex]Now, let's use the given information to express as inequality.
[tex]2(w+l)\leq56[/tex]However, we have to use another expression that relates the width and length.
[tex]l=w+4[/tex]Since the length is 4 units longer than the width. We replace this last expression in the inequality.
[tex]\begin{gathered} 2(w+w+4)\leq56 \\ 2(2w+4)\leq56 \\ 2w+4\leq\frac{56}{2} \\ 2w+4\leq28 \\ 2w\leq28-4 \\ 2w\leq24 \\ w\leq\frac{24}{2} \\ w\leq12 \end{gathered}[/tex]The largest width possible is 12 feet.
Now, we look for the length.
[tex]\begin{gathered} 2(12+l)\leq56 \\ 24+2l\leq56 \\ 2l\leq56-24 \\ 2l\leq32 \\ l\leq\frac{32}{2} \\ l\leq16 \end{gathered}[/tex]Therefore, the largest measure possible for the length is 16 feet.How to solve question 21? Area of the shaded region
The shaded region covers an area of 86.
Given that,
In the picture,
We have to find the area of the shaded region of question 21.
We know that,
The Area of the square is side square.
The area of the circle is πr².
The radius of the circle is 10.
We know that,
The circle's diameter is the same as a square's side length.
The diameter=10+10 =20
The side is 20
The area of the square
= 20² = 400
The area of the circle = π (10)²=π×100=314
Subtract the area of the square and area of the circle.
400-314
86
Therefore, The shaded region covers an area of 86.
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Linda must choose a number between 55 and 101 that is a multiple of 3,5, and 9. Write all the numbers that she could choose.
We will have the following:
*The LCM of the numbers given (3, 5 & 9) is 72. [This is the value to chosse]. We will have that 90 is also a multiple of 3, 5 & 9 [This is other value that can be chosen].
Here are the graphs of three equations:y = 50(1.5) ^xy = 50(2)^xY = 50(2. 5)^xWhich equation matches each graph? Explain how you know
The graphs below are exponential function graphs, the general formular takes the form
[tex]y=ab^x[/tex]The graph of
[tex]y=50(1.5)^x[/tex]Is shown below
The graph of
[tex]y=50(2^x)[/tex]Is shown below
The graph of
[tex]y=50(2.5^x)[/tex]Is shown below
Hence,
[tex]\begin{gathered} y=50(1.5)^x\rightarrow C \\ y=50(2)^x\rightarrow B \\ y=50(2.5)^x\rightarrow A \end{gathered}[/tex]The equation of the exponential function is
[tex]\begin{gathered} y=ab^x \\ a=50\rightarrow the\text{ initial value} \\ b\rightarrow growht\text{ factor} \end{gathered}[/tex]Thus the higher the growth factor the greater the rate of attaining a higher value within a short period.
That is why you see that the function with growth factor of 2.5 grows faster than that of 2 and also 1.5.
So the at x value of 3, the function with the greatest growth factor will have the highest y-value.
This implies , growth factor of 2.5 will have the highest, that corresponds to graph with colour green. Function with growth factor 2 will be the next to that of 2.5, that is red colored graph, and the last will be blue.
u ptsBirths are approximately Uniformly distributed between the 52 weeks of the year. They can be saidto follow a Uniform distribution from 1 to 53 (a spread of 52 weeks). Round answers to 4 decimalplaces when possible.a. The mean of this distribution isb. The standard deviation isC. The probability that a person will be born at the exact moment that week 18 begins isP(x = 18) =d. The probability that a person will be born between weeks 10 and 43 isP(10 < x < 43) =e. The probability that a person will be born after week 35 isP(x > 35)f. P(x > 18 x < 32) =g. Find the 47th percentile.h. Find the minimum for the upper quarter.
Step 1
A) The mean distribution
[tex]\frac{1+53}{2}=\frac{54}{2}=27.0000[/tex]Step 2
B) The standard deviation
[tex]\begin{gathered} SD=\sqrt[]{\frac{1}{12}\times(b-a)^2} \\ SD=\sqrt[]{\frac{1}{12}(53-1)^2} \\ SD=\text{ }15.0111 \end{gathered}[/tex]Step 3
C)
[tex]P(x=18)=0[/tex]Step 4
D)
[tex]\begin{gathered} P(10Step 5E)
[tex]P(x>35)=\text{ }\frac{53-35}{52}=\frac{18}{52}=0.3462[/tex]Step 6
F)
[tex]P(x>18|x<32)=\text{ }\frac{32-18}{32-1}=\frac{14}{31}=0.4516[/tex]Step 7
G)
[tex]\begin{gathered} \text{The 47th percentile}=1\text{ + }\frac{47}{100}(53-1)_{} \\ =1+0.47(52)=25.44_{}00 \end{gathered}[/tex]Step 8
[tex]\begin{gathered} \text{The minimum for the upper percentile = 1+((}\frac{3}{4})(53^{}-1) \\ =1+0.75(52) \\ =1+\text{ 39=40}.0000 \end{gathered}[/tex]find the second and third derivative of [tex]y = \sqrt{x} [/tex]
We can use the power rule to get the second and third derivative of the function.
[tex]\begin{gathered} \text{First derivative:} \\ y^{\prime}=\mleft(\frac{1}{2}\mright)x^{\frac{1}{2}-1} \\ y^{\prime}=(\frac{1}{2})x^{-\frac{1}{2}} \\ y^{\prime}=\frac{x^{-\frac{1}{2}}}{2} \end{gathered}[/tex][tex]\begin{gathered} \text{Second derivative} \\ y^{\prime}^{\prime}=(-\frac{1}{2})\frac{x^{-\frac{1}{2}-1}}{2} \\ y^{\prime\prime}=-\frac{x^{-\frac{3}{2}}}{4}\text{ or }y^{\prime\prime}=-\frac{1}{4x^{\frac{3}{2}}} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{Third derivative} \\ y^{\prime}^{\prime}^{\prime}=(-\frac{3}{2})-\frac{x^{-\frac{3}{2}-1}}{4} \\ y^{\prime\prime\prime}=\frac{3x^{-\frac{5}{2}}}{8}\text{ or }y^{\prime\prime\prime}=\frac{3}{8x^{\frac{5}{2}}} \end{gathered}[/tex]This expression represents the amount of money, in dollars, that will be in a savings account after 4 years.1500 [1 + 0.05/12)^12]^4Which of these is equivalent to the expression?A). 1500 (1.05/12)^48B). 1500 (1.05/12)^16C). 1500 (1 +0.05/12)^16D). 1500 (1 + 0.05/12)^48
The expression below represents the amount of money, in dollars, that will be in a savings account after 4 years.
[tex]1500\lbrack(1+\frac{0.05}{12})^{12}\rbrack^4[/tex]Recall from the laws of exponents, the power of a power rule is given by
[tex](a^x)^y=a^{x\cdot y}[/tex]So applying the above rule on the given expression, we get
[tex]\begin{gathered} 1500\lbrack(1+\frac{0.05}{12})^{12}\rbrack^4 \\ 1500(1+\frac{0.05}{12})^{12\cdot4} \\ 1500(1+\frac{0.05}{12})^{48} \end{gathered}[/tex]Therefore, the equivalent expression is
[tex]1500(1+\frac{0.05}{12})^{48}[/tex]Option D is the correct answer.
50th term 64 57 50 43...
hello
to solve this question, we need to know if this sequence is an arithmetic or geometric progression
first term (a) = 64
common difference (d) = -7
the nth term of an arithemetic progression is given as
[tex]\begin{gathered} T_n=a+(n-1)d_{} \\ n=\text{nth term} \\ a=\text{first term} \\ d=\text{common difference} \end{gathered}[/tex]now let's substitute the values into the equation above
[tex]\begin{gathered} T_n=a+(n-1)d_{} \\ a=64 \\ d=-7 \\ T_{50}=64+(50-1)\times-7 \\ T_{50}=64+(49\times-7) \\ T_{50}=64+(-343) \\ T_{50}=64-343 \\ T_{50}=-279 \end{gathered}[/tex]from the calculations above, the 50th term of the sequence is -279
not college I misclicked but the question is in pic
Answer
x = 13.33 units
Explanation
We can easily tell that the small triangle (with sides 6 and 8) is similar to the bigger triangle with sides (6+4 and x).
And the ratio of corresponding sides is the same for two similar triangles.
From the image, we can see that
6 is corresponding to (6 + 4)
8 is corresponding to x
So,
[tex]\begin{gathered} \frac{6}{6+4}=\frac{8}{x} \\ \frac{6}{10}=\frac{8}{x} \end{gathered}[/tex]We can now cross multiply
6x = (8) (10)
6x = 80
Divide both sides by 6
(6x/6) = (80/6)
x = 13.33 units
Hope this Helps!!!
Mrs. Davis has 20 people in her 6th period class. 12 of the people are boys. What percent of Mrs. Davis's 6th period class are boys? 70% 40% 50% 60%
Mrs. Davis has 20 people in her 6th period class. 12 of the people are boys. What percent of Mrs. Davis's 6th period class are boys? 70% 40% 50% 60%
we know that
20 people represent 100%
so
Applying proportion
20/100%=12/x
solve for x
x=(100*12)/20
x=60%
therefore
the answer is 60%Use the law of sines Find each missing side or angle
The law of sine states that the ratio of Sine A and side a is just equal to the ratio of Sine B and side b which is also equal to the ratio of Sine C and side c. In formula, we have:
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]where the big letter A, B, C are the angles and the small letters are the side opposite of the angle.
In our triangle, we have angle 19 and its opposite side is "x" whereas the angle opposite of the side that has a length of 32 units is unknown.
To solve the unknown angle, we know that the total measure of the angle in a triangle is 180 degrees. Therefore, the measure of the missing angle is 180 - 19 - 26 = 135 degrees.
So, going back to the law of sine, we have:
[tex]\begin{gathered} \frac{\sin19}{x}=\frac{\sin 135}{32} \\ \text{Cross multiply.} \\ 32\sin 19=x\sin 135 \\ \text{Divide both sides by sin 135.} \\ \frac{32\sin 19}{\sin 135}=\frac{x\sin 135}{\sin 135} \\ \frac{32\sin 19}{\sin 135}=x \\ \frac{10.41818094}{0.7071067812}=x \\ 14.73\approx x \end{gathered}[/tex]Therefore, the measure of the side x is approximately 14.73 units.
To solve the length of the other side, say y, the side opposite angle 26, we can make use of the law of sine again.
[tex]\begin{gathered} \frac{\sin135}{32}=\frac{\sin 26}{y} \\ y\sin 135=32\sin 26 \\ y=\frac{32\sin 26}{\sin 135} \\ y=\frac{14.0278767}{0.7071067812} \\ y\approx19.84 \end{gathered}[/tex]The length of the other missing side opposite angle 26 is approximately 19.84 units.
Answer:
14.7
Step-by-step explanation:
yes
At the sewing store, Kimi bought a bag of mixed buttons.The bag included 100 buttons, of which 10% were large.How many large buttons did kimi get?
to find the 10% of 100 buttons, we multiply 100 by 0.1 to get the following:
[tex]100\cdot0.1=10[/tex]therefore, Kimi got 10 large buttons
Pre-Calculus_Unit 1_Math_20-21 / 4 of 16 Find the slope of the line determined by the equation 3x +10y = 11 O A. m = -3 OB. m= 3 O C. 3 m=- 10 11 10 Em: -10
Brook, this is the solution:
Let's find the slope for this equation:
3x + 10y = 11
10y = -3x + 11
Dividing by 10 at both sides:
10y/10 = -3x/10 + 11/10
y = -3x/10 + 11/10
Therefore,
m = -3/10
For the interval expressed in the number line, write it using set-builder notation and interval notation.
Answer:
Writing the number line in set builder notation we have;
[tex]\mleft\lbrace x\mright|x>0\}[/tex]Writing in interval notation.
[tex]x=(0,\infty)[/tex]Explanation:
Given the number line in the attached image.
x starts on 0, with a non shaded circle and pointed to the right/positive direction.
So;
[tex]x>0[/tex]Writing the number line in set builder notation we have;
[tex]\mleft\lbrace x\mright|x>0\}[/tex]Writing in interval notation.
[tex]x=(0,\infty)[/tex]Since the upper boundary of x is not stated then we will represent it with infinity in the interval notation.
[tex]\begin{gathered} (\text{ }\rightarrow\text{ greater than} \\ \lbrack\text{ }\rightarrow\text{ greater than or equal to } \\ so,\text{ } \\ 0Find the area of the circle with a circumference of 62.8 inches. Use 3.14 for pi
The area of a circumference can be calculated with this formula:
[tex]C=2\pi r[/tex]Where "r" is the radius of the circle.
The area of a circle can be found with this formula:
[tex]A=\pi r^2[/tex]Where "r" is the radius of the circle.
If you solve for "r" from the formula of a circumference, you get:
[tex]r=\frac{C}{2\pi}[/tex]Knowing that:
[tex]\begin{gathered} C=62.8in \\ \pi\approx3.14 \end{gathered}[/tex]You get:
[tex]\begin{gathered} r=\frac{62.8in}{(2)(3.14)} \\ \\ r=10in \end{gathered}[/tex]Knowing the radius, you can find the area of the circle:
[tex]\begin{gathered} A=(3.14)(10in)^2 \\ A=314in^2 \end{gathered}[/tex]The answer is:
[tex]A=314in^2[/tex]I narrowly answered the first question on my homework but for some reason EF really confuses me.
Solution
Part 1
For this case we can find DF with the following proportion formula:
[tex]\frac{AC}{AB}=\frac{DF}{DE}[/tex]And replacing we got:
[tex]\frac{4}{2}=\frac{DF}{1.34},DF=2.68[/tex]Part 2
[tex]\frac{BC}{AB}=\frac{EF}{DE}[/tex]And solving for EF we got:
[tex]EF=1.34\cdot\frac{3}{2}=2.01[/tex]showing all your work for problem 1 divide simplify and state the domain and problem 2 multiply simplify and state the domain
The domain of the problem is given as
[tex](-\infty,\text{ 0) U (0, +}\infty)[/tex]A train leaves Little Rock, Arkansas, and travels North at 60 kilometers per hour. Another train leaves at the same time and travels South at 65 kilometers per hour. how many hours will it take before they are 250 kilometers apart?
After "t" seconds they will be
65*t + 60*t seconds apart, therefore we are looking for a "t" such that
65*t + 60*t = 250
125*t = 250
so t=2
A local bakery, theprice for abughouts for his employeespurchasedAFX) -0.65- 3.5prie for $3.50 and some doudoughnuts 30.05. Each day the manager at the store buyswhich equation represents the total cosa function of the number of doughtswhich equation on where represents the number of tires produced over resismodels the function ?BFX) - 0.65x + 3.5CX-3.5x + 0.85DX) - 3.5x -0.85Aebire manufacturing plant produces soo tires a day on average. If the production ofAFX) - 500 + xB (x) - 500 -Cx) - 500xDX) - 5006.Bushra purchases a car for $12,900. The car will depreciate at a rate of 15% each year,After how many years will the value of the car bethan $3,000?A 6 yearsB 7 yearsC8 yearsD 9 years
In order to create a function that represents the cost of the manager as a function of the number of doughnuts he buys, we need to multiply the cost of each doughnut ($ 0.85) by the number of employees the manager has and add the value of the pie ($ 3.5). This is done below:
[tex]f(x)\text{ = }0.85\cdot x\text{ + 3.5}[/tex]The correct option is the letter B.
The car starts at $ 12,900 and depreciate at a rate of 15% each year. This means that the value of the car on any given year is ruled by the tollowing expression:
[tex]M\text{ = C}\cdot(1-r)^t[/tex]Where "M" is the value of the car after "t" years, C is the initial value of the car and "r" is the rate at which the car depreciates every year divided by 100. Aplying the data from the problem on the expression gives us:
[tex]3000\text{ = 12900}\cdot(1\text{ - }\frac{15}{100})^t[/tex]We want to solve for the variable "t", because we want to know how many years it'll take until the car reaches the final value of 3000.
[tex]\begin{gathered} 12900\cdot(1\text{ - }\frac{15}{100})^t\text{ = 3000} \\ (1\text{ - }\frac{15}{100})^t\text{ = }\frac{3000}{12900} \\ (1-0.15)^t\text{ = }\frac{30}{129} \\ (0.85)^t\text{ = }\frac{30}{129} \end{gathered}[/tex]We have reached an exponential equation. To solve it we need to aply a logarithm on both sides of the equation.
[tex]\begin{gathered} \ln (0.85^t)\text{ = }\ln (\frac{30}{129}) \\ t\cdot\ln (0.85)\text{ = }ln(30)\text{ - ln(129)} \\ t\cdot(-0.1625)\text{ = }3.4\text{ - 4.86} \\ t\text{ = }\frac{-1.46}{-0.1625}\text{ = 8.98} \end{gathered}[/tex]It'll take approximately 9 years to reach that value. The correct option is the letter "D".
If x is multiplied by 5 and then 3 is subtracted, then the function isf(x) = 5x -3.What are the steps to find the inverse to this function?
Step:
Concept:
First, find the inverse of subtraction which is addition
x + 3
Step 2:
The multiplicative inverse is division, hence, you will divide x + 3 by 5.
Therefore, we have
[tex]\begin{gathered} y\text{ = }\frac{x\text{ + 3}}{5} \\ \end{gathered}[/tex]The inverse of the function is given below.
[tex]f^{-1}(x)\text{ = }\frac{x\text{ + 3}}{5}[/tex]Method 2
[tex]\begin{gathered} \text{If f(x) = 5x - 3} \\ \text{let y = 5x - 3} \\ \text{Make x subject of the formula} \\ \text{y + 3 = 5x} \\ x\text{ = }\frac{y\text{ + 3}}{5} \\ \text{Write the inverse of f(x) by changing y to x} \\ f^{-1}(x)\text{ = }\frac{x\text{ + 3}}{5} \end{gathered}[/tex]Answer:
Add 3, then divide by 5
Step-by-step explanation:
help please and thankyou Write an equation that represents the graph of the line shown in the coordinate plane below.
Slope intercept-form equation:
y=mx+b
Where:
m= slope
b= y-intercept
by looking at the graph we can see that the line crosses the origin (0,0), so it crosses the y-axis at x=0.
b=0
For the slope:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]point 1= (x1,y1)= (-2,-4)
Point2= (x2,y2)= (4,8)
Replacing:
[tex]m=\frac{8-(-4)}{4-(-2)}=\frac{12}{6}=2[/tex]So the final expression:
y= 2x
A tourist at scenic Point Loma, California uses a telescope to track a boat approaching the shore. If the boat moves at a rate of5 meters per second, and the lens of the telescope is 30 meters above water level, how fast is the angle of depression of thetelescope (0) changing when the boat is 200 meters from shore? Round any intermediate calculations to no less than sixdecimal places, and round your final answer to four decimal places.
Lest first we hte sine theorem to relate the given measures:
[tex]\frac{\sin (\theta)}{30}=\frac{\sin(90^{\circ})}{\sqrt[]{x^2+30^2}}[/tex]x represents the distance from the boat to the shore.
[tex]\frac{\sin (\theta)}{30}=\frac{1}{\sqrt[]{x^2+30^2}}[/tex][tex]\frac{\sin(\theta)}{1}=\frac{30}{\sqrt[]{x^2+30^2}}[/tex][tex]\sin (\theta)=\frac{30}{\sqrt[]{x^2+30^2}}[/tex][tex]\theta=\sin ^{-1}(\frac{30}{\sqrt[]{x^2+30^2}})[/tex]Then we must calculate the derivative in order to know the rate of change at a certain point.
[tex]\frac{d}{dx}(\sin ^{-1}(\frac{30}{\sqrt[]{x^2+30^2}}))=-\frac{30x}{\sqrt[]{\frac{x^2}{x^2+900}}\cdot(x^2+900)^{\frac{3}{2}}}[/tex]To find how fast is the angle of depression of the telescope is changing when the boat is 200 meters from shore, replace by 200 on the derivative:
[tex]-\frac{30\cdot200}{\sqrt[]{\frac{200^2}{200^2^{}+900}}\cdot(200^2+900)^{\frac{3}{2}}}=-0.0007\text{ rad/s}[/tex]Which expression is equivalent to 8c + 6 - 3c - 2 ?A. 5c +4B. 50 + 8C.11c +4D.11c + 8
This means that the answer is option A
What the percent 7/800
You have to divide 7 by 800:
[tex]\frac{7}{800}=0.00875[/tex]Now multiply by 100
[tex]0.00875*100=0.875\%[/tex]The answer is 0.875%
35/25 covert fraction to percent
To convert fraction to decimal you multiply by 100%
Therefore, 35/25 to percentage
[tex]\begin{gathered} =\text{ }\frac{35}{25}\text{ x 100\%} \\ =\text{ }\frac{35\text{ x 100}}{25} \\ =\text{ }\frac{3500}{25} \\ =\text{ 140\%} \end{gathered}[/tex]Brenda Ortiz earns $18,200 per year. Find her semimonthly salary
Semi monthly salary are paid twice a month, since the year has twelve months then the total number of semi monthly salaries she'll recieve is:
[tex]\text{n = 12}\cdot2\text{ = 24}[/tex]She'll receive 24 salaries in a year. To calculate the value of each salary we need to divide the total amount she earns in a year by 24.
[tex]\text{semimonthly = }\frac{18200}{24}\text{ = }758.34[/tex]Her semi monthly salary is $758.34.
A researcher would like to determine whether a new drug has an effect on IQ. A sample of n = 100 participants is obtained, and each person is given a standard dose of the medication one hour beforebeing given an IQ test. For the general population, scores on the IQ test are normally distributed with μ= 100 and o=15. The individuals in the sample who took the drug had an average score of M =103.a. Use a two-tailed test with a= .05. Conduct the four steps for hypothesis testing and labeleach step: Step1, Step 2, Step 3, and Step 4.b. Calculate Cohen's d.c. Are the data sufficient to conclude that there is a significant difference? Write youranswer in the form of a sentence.
There is sufficient evidence to conclude that there is a significant difference in the hypothesis testing of the normal distribution.
a) Let us consider the hypothesis for the experiment.
Step 1:
H₀ : μ = 100
Similarly now we find H₁
H₁ : μ ≠ 100
Step 2:
Now we have to find the test statistic.
Z = (x - μ )/(σ÷√n)
Now we have x =103 , μ = 100 , σ =15 and n = 100
Z= (103-100) /15 × 10
Z=2.00
Step 3:
Now critical region is rejected if Z > Z(α/2)
Z(α/2) = 1.96
Now H₀ is rejected if z >-1.96 nor Z>1.96
Step 4:
As Z= 2>1.96 hence we will reject the null hypothesis.
b) Cohen's dc = 103 -100 / 15 = 0.2
c) Hence we can say that there is sufficient evidence to conclude that there is a significant difference.
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4. Tickets for a carnival cost $6 for adults and $4 for children. The school has abudget of $120 for a field trip to the carnival. An equation representing thebudget for the trip is 120 = 6x + 4y. Here is a graph of this equation:
Given:
The equation is 6x + 4y = 120.
Explanation:
The points that lies on the line satifies the equation. So point (0,30) lies on the number which 0 adults and 30 children could go to school. So "if no adult chaperons were needed, 30 students could go to school is true.
For ten students and 15 adults point is (15,10). The point (15,10) does not lie on number line and not satifies the equation so second statement is false.
The cost of tickets for 4 adults is,
[tex]4\cdot6=24[/tex]and cost of tickets for six students is,
[tex]6\cdot4=24[/tex]Both costs are equal, means for six fewer students 4 additional adults can go to the zoo. Thus third statement is correct.
The cost of tickets for two children is,
[tex]4\cdot2=8[/tex]The cost of tickets for 3 adults is,
[tex]6\cdot3=18[/tex]Since cost of tickets for 3 adults is more than cost of tickets for two children which means two children can not go to the zoo for 3 fewer adults in the trip. Thus fourth statement is wrong.
For 16 adults and 6 students point is (16,6). The point (16,6) lies on the number line, which point (16,6) satifies the equation. So fifth statement is correct.