I need help with this math homework!!!

A student would like to find the height of a statue. The length of the statue's right arm is 54 feet. The student's right arm

Komok [63]

Answer:

a) Approximate height of the statue: 144 feet

b) The approximate height is 1.7% greater that the actual height of the statue.

Explanation

1) Assume that the measures of the statue are proportional to the dimensions of the student. That means:

statue's height student's height

________________________ = ______________________

length of the statue's right arm length of student's right arm

⇒ x / 54 feet = (5 + 1/3) feet / 2 feet

⇒ x = 54 × (16/3) / 2 feet = (54×16) / (3×2) feet = 144 feet

Answer: 144 feet

2) Compare the approximate heigth obtained with the actual height:

144 feet - (143 feet + 9 inches) = 144 feet - (143 feet + 9/12 feet) = 1 feet - 9/12 feet

1 feet - 9/12 feet = 1 feet - 3/4 feet = 1/4 feet.

Hence, the approximate feet obtained is 1/4 feet larger than the actual height.

In relative terms that is : (1/4) / (143 + 3/4) = 0.0017 = 1.7%.

4 0

3 years ago

Please help its easy

Ainat [17]

Answer:

$\frac{3}{4}$

Step-by-step explanation:

$\frac{12 - 3}{12}$

$\frac{9}{3}$

cross out the common factor.

$\frac{3}{4}$

8 0

3 years ago

Read 2 more answers

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

Which expressions can never result in a negative real number when evaluated for any value of x? Selecta

Y_Kistochka [10]

I think it’s is 5 and take it’s now

8 0

3 years ago

this month salespersons a made 11% of $67,530. salesperson b made 8% of $85,740. who made more commission this month? how much d

Paul [167]

Salesperson A made $7428.30
Salesperson B made $6859.20
Salesperson A made more commission this month and made $7428.30

7 0

4 years ago

Read 2 more answers