Find the missing side or angle.
Round to the nearest tenth.

Answer:

The probability is 0.31084

Step-by-step explanation:

We can calculate this probability using the z-score route.

Mathematically;

z = (x-mean)/SD/√n

Where the mean = 16, SD = 3 and n = 36

For 15.8, we have;

z = (15.8-16)/3/√36 = -0.2/3/6 = -0.2/0.5 = -0.4

For 16.2, we have

z = (16.2-16)/3/√36 = 0.2/3/6 = 0.2/0.5 = 0.4

So the probability we want to calculate is;

P(-0.4<z<0.4)

We can get this using the standard normal distribution table;

So we have;

P(-0.4 <z<0.4) = P(z<-0.4) - P(z<0.4)

= 0.31084

Answer:

(-2,-7)

Step-by-step explanation:

because it's reflected across the x-axis, only the y-intercept will change

Answer:

(-2,-7)

Step-by-step explanation:

All you have to do is draw a graph and draw the point across the x axis in the same row and same distance from the x axis.The distance is 7 so you just change it to -7.

Answer:

Step-by-step explanation:

[tex](1+6i)-(7+3i) ?\\Group\:the\:real\:part\:and\:the\:imaginary\\\:part\:of\:the\:complex\:number\\\left(a+bi\right)\pm \left(c+di\right)=\left(a\:\pm \:c\right)+\left(b\:\pm \:d\right)i\\=\left(1-7\right)+\left(6-3\right)i\\1-7=-6\\6-3=3\\=-6+3i[/tex]

Answer:

150000 doctoral degrees

Step-by-step explanation:

From the graph attached,

At t = 0 represents the year 2005 and each unit of y-axis represents 10000 degrees.

Number of approximate number of degrees awarded to women in 2008,

f'(3) = 2.25

Similarly, number of doctoral degrees awarded from 2009 to 2013 are,

f'(4) = 2.5

f'(5) = 2.5

f'(6) = 2.5

f'(7) = 2.5

f'(8) = 2.75

Total number of degrees awarded to women from the start of 2008 to the start of 2014 = (2.25 + 2.5 + 2.5 + 2.5 + 2.5 + 2.75 ) × 10000

= 15 × 10000

= 150000

Answer: C. 41

Step-by-step explanation:

[tex]-\left(31+2\right)+7^2-\left(-5^2\right)[/tex]

[tex]=-33+7^2-\left(-5^2\right)[/tex]

[tex]\left(-5^2\right)=-25[/tex]

[tex]=-33+7^2-\left(-25\right)[/tex]

[tex]7^2=49[/tex]

[tex]=-33+49-\left(-25\right)[/tex]

[tex]-33+49=16[/tex]

[tex]=16-\left(-25\right)[/tex]

[tex]\mathrm{Apply\:rule\:}-\left(-a\right)\:=\:+a[/tex]

[tex]16+25=41[/tex]

Answer:

3 ( a ) : x = 3.6,

3 ( b ) : x = 5

Step-by-step explanation:

For 3a, we can calculate the value of x through Pythagorean Theorem, which seemingly was your approach. However, the right triangle with x present as the leg, did not have respective lengths 9.6 and 12. The right angle divides 9.6 into two congruent parts, making one of the legs of this right triangle 9.6 / 2 = 4.8. The hypotenuse will be 12 / 2 as well - as this hypotenuse is the radius, half of the diameter. Note that 12 / 2 = 6.

( 4.8 )² + x² = ( 6 )²,

23.04 + x² = 36,

x² = 36 - 23.04 = 12.96,

x = √12.96, x = 3.6

Now as you can see for part b, x is present as the radius. Length 3 forms a right angle with length 8, dividing 8 into two congruent parts, each of length 4. We can form a right triangle with the legs being 4 and 3, the hypotenuse the radius. Remember that all radii are congruent, and therefore x will be the value of this hypotenuse / radius.

( 4 )² + ( 3 )² = ( x )²,

16 + 9 = x² = 25,

x = √25, x = 5

Answer:

Step-by-step explanation:

I'm having difficulty reading your input:  25, y, + 18, 25.  Please clarify your meaning.

5x3 – 6x2 is a polynomial expression with two terms.

The term  is a ratio.  False.   See above.

The entire expression is a difference.  True.

There are four terms.  False.   See above.

Two statements B and C are true for the expression 5x3 – 6x2 25y+ 18

A. The term is a ratio. False

B. There are three terms.terms. True

C. The entire expression is a difference. True

D. There are four terms. False

Learn more about linear equation here: brainly.com/question/1884491

#SPJ2

Answer:

Step-by-step explanation:

Given that:

Mean = 3.34

sample size = 26

sample mean = 2.7

standard deviation = 1.17

level of significance = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu \geq 3.34} \\ \\ \mathtt{H_1: \mu < 3.34}[/tex]

degree of freedom = n - 1

degree of freedom = 26 -1

degree of freedom =  25

level of significance = 0.10

Since the alternative hypothesis contains <, then the test is left tailed

[tex]\mathtt{t_{\alpha, df} = t_{0.10, 25}}[/tex]

[tex]\mathtt{t_{0.10, 25}}[/tex] = - 1.316

The rejection region therefore consist of all values smaller than - 1.316, therefore ; reject [tex]H_o[/tex] if t < -1.316

The test statistics can be computed as follows:

[tex]t = \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]t = \dfrac{2.7 - 3.34}{\dfrac{1.17}{\sqrt{26}}}[/tex]

[tex]t = \dfrac{-0.64}{\dfrac{1.17}{5.099}}[/tex]

t = - 2.789

Decision Rule:  To reject the null hypothesis if the t test lies in the rejection region or less than the rejection region.

Conclusion: We  reject the null hypothesis since t = (- 2.789) < -1.316. Then we conclude that  the mean number of residents in the retirement community household is less than 3.34 persons.

Answer:

add 8x to both sides

Step-by-step explanation:

5-8x<2x+3

first step, subtract 3 from both sides:

2-8x<2x

second step,?

2<?x

so you need to add 8x first

Answer:

6

Step-by-step explanation:

Hello, please consider the following.

[tex](4+x)^2=4^2+2\cdot 4\cdot x+x^2=16+\boxed{8}x+x^2\\\\\text{ ... and not ...}\\\\16+\boxed{4}x+x^2[/tex]

So the correct equation becomes.

[tex]x^2+64=16+8x+x^2\\\\8x=64-16=48\\\\\text{ we divide by 8 both sides of the equation.}\\\\x=\dfrac{45}{8}=6[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

Error : The expression ( 4 + x )² was expanded incorrectly.

Correct Solution : x = 6

Step-by-step explanation:

The planning of the solution is correct, by Pythagorean Theorem you can say that PQ² + QO² = PO², and hence through substitution x² + 8² = ( 4 + x )². Let's look into the calculations.

PQ² + QO² = PO²,

x² + 8² = ( 4 + x )²,

x² + 8² = 16 + 8x + x²,

64 = 16 + 8x,

48 = 8x,

x = 48 / 8 = 6, x = 6

As you can see, the only error in the calculations was expanding the expression ( 4 + x )². ( 4 + x )² = 4² + 2 [tex]*[/tex] 4 [tex]*[/tex] x + x² = 4² + 8x + x² = 16 + 8x + x², not 16 + 4x + x².

Answer:

155 + 165.3 - 65.5 - 23.25 - 26.45

Step-by-step explanation:

She had $155 dollars in the starting = +155

She withdrew $65.5 = -65.5

She withdrew another $23.25 = -23.25

She withdrew another $26.45 = -26.45

The deposited $165.3 = +165.3

The expression looks like:

155 + 165.3 - 65.5 - 23.25 - 26.45

We could simplify the expression:

155 + 165.3 - 65.5 - 23.25 - 26.45

=> 320.3 - 88.75 - 26.45

=> 320.3 -115.2

=> 205.1

At the end of the week, she had a total of $205.10.

Answer:

84 pounds

Step-by-step explanation:

If 1/3 of a book is equal to 28 pounds then 28*3 will give you your answer

False.

2x10^6 you move the decimal point 6 places to the right. ( add 6 zeros after the 2)

2x 10^6 = 2,000,000

Answer:

AB = 11

Step-by-step explanation:

Using the external secant property,

CB*CA = CD*CL (=CT^2  if T is tangent throught C)

substitute values,

7*(7+2x+3) = 9*(9+x+1)

Expand and simplify

70+14x = 90 + 9x

5x = 20

x = 4

=>

AB = 2x+3 = 2(4)+3 = 8+2 = 11

Answer:

The percentage increase in the production cost of the printer is 3%.

Step-by-step explanation:

We are given that the production of a printer consists of the cost of raw material at 100 dollars the cost of overheads at 80$ and wages at 120$.

Also, the cost of raw materials and overheads are increased by 11% and 20% respectively while wages are decreased by 15%.

Cost of raw material = $100

Cost of overheads = $80

Cost of wages = $120

So, the total cost of the printer = $100 + $80 + $120

                                                   = $300

Now, the increase in the cost of raw material = $100 + 11% of $100

                                                                           = [tex]\$100 + (\frac{11}{100} \times \$100)[/tex]

                                                                           = $100 + $11 = $111

The increase in the cost of overheads = $80 + 20% of $80

                                                                = [tex]\$80 + (\frac{20}{100} \times \$80)[/tex]

                                                                = $80 + $16 = $96

The decrease in the cost of wages = $120 - 15% of $120

                                                          = [tex]\$120 - (\frac{15}{100} \times \$120)[/tex]

                                                          = $120 - $18 = $102

So, the new cost of a printer = $111 + $96 + $102 = $309

Now, the percentage increase in the production cost of the printer is given by;

      % increase =  [tex]\frac{\text{Net increase in the cost of printer}}{\text{Original cost of printer}} \times 100[/tex]

                         =  [tex]\frac{\$309- \$300}{\$300} \times 100[/tex]

                         =  3%

Hence, the percentage increase in the production cost of the printer is 3%.

Answer:

a

   The  90%  confidence interval is  [tex]52561.13 < \mu < 57540.8[/tex]

b

Confidence interval for the population men between $52561.13  up to $57540.8

Step-by-step explanation:

From the question we are told that

   The sample size is  [tex]n = 25[/tex]

     The  sample mean is  [tex]\= x = \$ 55,051[/tex]

     The standard deviation is  [tex]\sigma = \$ 7,568[/tex]

Given that the confidence level is  90% then the level of confidence is mathematically represented as

             [tex]\alpha = 100 -90[/tex]

              [tex]\alpha = 10\%[/tex]

             [tex]\alpha = 0.10[/tex]

Next we obtain the critical value of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table the values is  

               [tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]

Generally the margin of error is mathematically  represented as

               [tex]E = Z_{\frac{ \alpha }{2} } * \frac{ \sigma }{\sqrt{n} }[/tex]

substituting values

               [tex]E = 1.645 * \frac{ 7568}{ \sqrt{ 25} }[/tex]

             [tex]E = 2489.9[/tex]

The 90% confidence interval is mathematically evaluated as

       [tex]\= x -E < \mu < \= x +E[/tex]

substituting values

     [tex]55051 - 2489.8 < \mu < 55051 + 2489.8[/tex]

     [tex]52561.13 < \mu < 57540.8[/tex]

Answer:

Shift the graph of y = x2 right 6 units.

Answer:

A; The first choice.

Step-by-step explanation:

We have the equation [tex]x^4+6x^2+5=0[/tex] and we want to solve using u-substitution.

When solving by u-substitution, we essentially want to turn our equation into quadratic form.

So, let [tex]u=x^2[/tex]. We can rewrite our equation as:

[tex](x^2)^2+6(x^2)+5=0[/tex]

Substitute:

[tex]u^2+6u+5=0[/tex]

Solve. We can factor:

[tex](u+5)(u+1)=0[/tex]

Zero Product Property:

[tex]u+5=0\text{ and } u+1=0[/tex]

Solve for each case:

[tex]u=-5\text{ and } u=-1[/tex]

Substitute back u:

[tex]x^2=-5\text{ and } x^2=-1[/tex]

Take the square root of both sides for each case. Since we are taking an even root, we need plus-minus. Thus:

[tex]x=\pm\sqrt{-5}\text{ and } x=\pm\sqrt{-1}[/tex]

Simplify:

[tex]x=\pm i\sqrt{5}\text{ and } x=\pm i[/tex]

Our answer is A.

Answer:

The acceleration on the unit is 0.667 m/s^2

The tension on the draw-bar is 13.34 kN

Step-by-step explanation:

The mass of the truck = 10 Mg = 10 x 10^3 kg

The mass of the trailer = 20 Mg = 20 x 10^3 kg

Tractive force from the truck = 20 kN = 20 x 10^3 N

The total mass of the unit = 10 Mg + 20 Mg = 30 Mg = 30 x 10^3 kg

The tractive force on the unit will produce an acceleration that is given as

F = ma

where

F is the tractive = 20 x 10^3 N

m is the mass of the unit = 30 x 10^3 kg

a is the acceleration of the unit = ?

substituting into the equation

20 x 10^3 = 30 x 10^3 x a

a = (20 x 10^3)/(30 x 10^3) = 0.667 m/s^2

the tension on the draw-bar T is gotten from considering only the mass that is pulled by the draw-bar which is 20 Mg

The acceleration on the unit = 0.667 m/s^2

The drawn mass = 20 Mg = 20 x 10^3 kg

The tension on the draw bar = ma = 20 x 10^3 x 0.667 = 13340 N

= 13.34 kN

The acceleration is 0.00067m/s^2, while the tension on the horizontal bar is 13.4 N

The given parameters are:

[tex]\mathbf{m = 10Mg}[/tex] -- mass of the truck

[tex]\mathbf{M = 20Mg}[/tex] -- mass of the trailer

[tex]\mathbf{F_T = 20kN}[/tex] --- tractive force

Start by calculating the total mass

[tex]\mathbf{M_T = m + M}[/tex]

So, we have:

[tex]\mathbf{M_T = 10Mg + 20Mg}[/tex]

[tex]\mathbf{M_T = 30Mg}[/tex]

Convert to kilograms

[tex]\mathbf{M_T = 30 \times 10^3kg}[/tex]

[tex]\mathbf{M_T = 30000 kg}[/tex]

Force is calculated as:

[tex]\mathbf{F =ma}[/tex]

So, we have:

[tex]\mathbf{20kN =30000kg \times a}[/tex]

Divide both sides by 30000

[tex]\mathbf{a = 0.00067ms^{-2}}[/tex]

The tension on the horizontal bar (i.e. the 20 Mg trailer) is:

[tex]\mathbf{T=ma}[/tex]

So, we have:

[tex]\mathbf{T=20Mg \times 0.00067ms^{-2}}[/tex]

Rewrite as:

[tex]\mathbf{T=20 \times 10^3 kg \times 0.00067m/s}[/tex]

[tex]\mathbf{T=13.4N}[/tex]

Hence, the acceleration is 0.00067m/s^2, while the tension on the horizontal bar is 13.4 N

Read more about force and acceleration at:

https://brainly.com/question/20511022

Answer:

(3x+2)^2

Step-by-step explanation:

Answer:

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Step-by-step explanation:

Given

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100})[/tex]

Required

Simplify

For clarity, group the expression in threes

[tex]((\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the first group [Divide 8 by 4]

[tex]((\frac{3}{1})*(\frac{2}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 9 by 3]

[tex]((\frac{1}{1})*(\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex]((\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 15 by 3]

[tex]((\frac{2}{1})*(\frac{5}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 16 by 2]

[tex]((\frac{1}{1})*(\frac{5}{8}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the second group [Divide 35 and 25 by 5]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{7}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 49 by 7]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{1}{3})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 24 by 3]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{1}{1})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Merge the first and second group

[tex]((\frac{5}{8})*(\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](1*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the last group [Divide 99 by 9]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{9})*(\frac{11}{100}))[/tex]

[Divide 63 by 9]

[tex](\frac{4}{7})*((\frac{7}{64})*(\frac{80}{1})*(\frac{11}{100}))[/tex]

[Divide 64 and 80 by 8]

[tex](\frac{4}{7})*((\frac{7}{8})*(\frac{10}{1})*(\frac{11}{100}))[/tex]

[Divide 10 and 4 by 2]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{5}{1})*(\frac{11}{100}))[/tex]

[Divide 100 by 5]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{1}{1})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*(\frac{7}{4})*(\frac{11}{20})[/tex]

[tex]1*(\frac{11}{20})[/tex]

[tex]\frac{11}{20}[/tex]

Hence;

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Answer:

1 hour

Step-by-step explanation:

Hello, let's say that her departure trip takes t in minutes, as her return speed is 3 times her departure speed, she took t/3 for the return and we know that this 40 minutes less, so we can write.

t/3=t-40

We can multiply by 3

t = 3t -40*3 = 3t - 120

This is equivalent to

3t -120 = t

We subtract t

2t-120 = 0

2t = 120

We divide by 2

t = 120/2 = 60

So this is 60 minutes = 1 hour.

Thank you.

Answer:

[tex]\Large \boxed{\sf \bf \ \ y=\dfrac{1}{16}(a-3)^2-2 \ \ }[/tex]

Step-by-step explanation:

Hello, please consider the following.

When the parabola equation is like

[tex]y=a(x-h)^2+k[/tex]

The vertex is the point (h,k) and the focus is the point (h, k+1/(4a))

As the vertex is (3,-2) we can say that h = 3 and k = -2.

We need to find a.

The focus  is (3,2) so we can say.

[tex]2=-2+\dfrac{1}{4a}\\\\\text{*** We add 2. ***}\\\\\dfrac{1}{4a}=2+2=4\\\\\text{*** We multiply by 4a. ***}\\\\16a=1\\\\\text{*** We divide by 16. ***}\\\\a=\dfrac{1}{16}[/tex]

So an equation for the parabola is.

[tex]\large \boxed{\sf y=\dfrac{1}{16}(a-3)^2-2 }[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

Step-by-step explanation:

Given that :

the null and the alternative hypothesis are computed as :

[tex]H_o: \mu_1 = \mu_2[/tex]

[tex]H_a : \mu _1 \neq \mu_2[/tex]

This is a two tailed test

This is because of the ≠ sign in the alternative hypothesis which signifies that the rejection region in the alternative hypothesis are at the both sides of the hypothesized mean difference .

Decision Rule: at the level of significance ∝ = 0 . 10

The decision rule is to reject the null hypothesis if z < - 1 . 64 and z > 1 . 64

NOTE: DURING THE MOMENT OF TYPING THIS ANSWER THERE IS A TECHNICAL ISSUE WHICH MAKES ME TO BE UNABLE TO SUBMIT THE FULL ANSWER BUT I'VE MADE SCREENSHOTS OF THEM AND THEY CAN BE FOUND IN THE ATTACHED FILE BELOW

Answer:

- At any time t, the population is:

P = 375t² + 3000t + 1000

- At time t = 3 days, the population is:

P = 13,375

Step-by-step explanation:

Given the rate of change of the population of bacteria as:

dP/dt = 3000/(1 + 0.25t)

we need to rewrite the given differential equation, and solve.

Rewriting, we have:

dP/3000 = (1 + 0.25t)dt

Integrating both sides, we have

P/3000 = t + (0.25/2)t² + C

P/3000 = t + 0.125t² + C

When t = 0, P = 1000

So,

1000/3000 = C

C = 1/3

Therefore, at any time t, the population is:

P/3000 = 0.125t² + t + 1/3

P = 375t² + 3000t + 1000

At time t = 3 days, the population is :

P = 375(3²) + 3000(3) + 1000

= 3375 + 9000 + 1000

P = 13,375

Midpoint formula: (x1 + x2)/2 , (y1 + y2)/2

Midpoint AB = (3 +-5)/2, (3 + 7)/2 = -2/2 , 10/2 = (-1,5)

Midpoint CD = (2 +9)/2, (11 + -2)/2 = (11/2,9/2)

Answer:

34 grams

Step-by-step explanation:

If the remaining sample has 30.26 grams of radioactive substance, and 11% of it decayed, that means that 30.26 grams is 89% of the original. Let the original be x.

30.26=0.89x

Multiply both by one hundred

3026=89x

Divide both by 89

34=x

x=original, so the original was 34 grams.

Answer:

Increasing

0°≤x≤180°

Decreasing

180°≤x≤360°

Answer:

Step-by-step explanation:

Formula for area of a triangle:

Height x Base /2

Base (b) = h +4

Height = h

h + 4 x h /2 = 30cm

=> h +4 x h = 60

=> h+4h =60

=> 5h = 60

=> h = 12

Height = 12

Base = 12 +4 = 16

Answer:

Option A (repeated measures design) is the correct option.

Step-by-step explanation:

The other three options are not related to the given instance. So that alternative A would be the correct choice.