Evaluate the following expression:<br> 25x(x

In Mrs. Smith’s enrichment class, 25% of the students are boys. There are 7 boys in the class. How many total students are in Mr

Yakvenalex [24]

Answer:28

Step-by-step explanation: 25% is 1/4 so you can multiply 7 and 4 to get 28

4 0

3 years ago

A rectangular parcel of land has an area of 8,000 ft2. a diagonal between opposite corners is measured to be 10 ft longer than o

lapo4ka [179]

Answer:

Step-by-step explanation:

Let the area of the rectangle be Length × width

LW = 8000

If the diagonal is 10feer greater than one of its length then:

D = 10+L

Diagonal is represented using the Pythagoras theorem as shown:

D² = L²+W²

(10+L)²= L²+W²

100+20L +L² = L²+W²

From LW = 8000

W = 8000/L

100 + 20L = (8000/L)²

100 +20L = 64,000,000/L²

Cross multiply

100L² + 20L³ = 64,000,000

2L³+10L² = 6,400,000

L³+5L² = 3200000

L³+5L² -3200000 = 0

5 0

4 years ago

Read 2 more answers

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 229 wi

meriva

Answer:

A. 99% CI for p: 0.125 < p < 0.259

B. 99% CI for p: 0.807 < p < 0.911

C. 95% CI for p: 0.776 < p < 0.844

D. 95% CI using t-distribution: 23.7 < µ < 40.3

Step-by-step explanation:

A.

$p=\frac{x}{n} =\frac{44}{229} = 0.192139738$

Confidence = 99% = 0.99

$\alpha$ =1 - confidence = 1 - 0.99 = 0.01

Critical z-value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.192139738(1-0.192139738)}{229} }$

$SE=\sqrt{\frac{0.155222059}{229} }$

SE = 0.026035

E = Margin of Error

E = z-value * SE = 2.58 * 0.026035

E = 0.067170516

Lower Limit = p - E = 0.192139738 - 0.067170516 = 0.125

Upper Limit = p + E = 0.192139738 + 0.067170516 = 0.259

99% CI for p: 0.125 < p < 0.259

B.

$p=\frac{x}{n} =\frac{256}{298} = 0.8590604$

Confidence = 99% = 0.99

$\alpha$ =1 - confidence = 1 - 0.99 = 0.01

Critical z-value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.8590604(1-0.8590604)}{298} }$

$SE=\sqrt{\frac{0.121075629}{298} }$

SE = 0.020157

E = Margin of Error

E = z-value * SE = 2.58 * 0.020157

E = 0.052004

Lower Limit = p - E = 0.8590604 - 0.052004 = 0.807

Upper Limit = p + E = 0.8590604 + 0.052004 = 0.911

99% CI for p: 0.125 < p < 0.259

C.

$p=\frac{x}{n} =\frac{405}{500} = 0.81$

Confidence = 95% = 0.95

$\alpha$ =1 - confidence = 1 - 0.95 = 0.05

Critical z-value = $z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.81(1-0.81)}{500} }$

$SE=\sqrt{\frac{0.1539}{500} }$

SE = 0.017544

E = Margin of Error

E = z-value * SE = 1.96 * 0.017544

E = 0.034387

Lower Limit = p - E = 0.81 - 0.034387 = 0.776

Upper Limit = p + E = 0.81 + 0.034387 = 0.844

95% CI for p: 0.776 < p < 0.844

D.

x = sample mean = 32

s = sample standard deviation = 13

n = sample size = 12

Confidence = 95% = 0.95

$\alpha$ =1 - confidence = 1 - 0.95 = 0.05

Being the sample size less than 30, we use the statistical t.

df = degree of freedom = n - 1 = 12 - 1 = 11

Critical t-value = $t_{\alpha/2,df}=t_{0.05/2,11}=t_{0.025,11}=2.201$ (from t-table, two tails, df=11)

SE = standard error

$SE=\frac{s_{x} }{\sqrt{n} } =\frac{13 }{\sqrt{12} }=3.752777$

E = margin of error

E = t-value * SE = 2.201 * 3.752777 = 8.259862

Lower Limit = x - E = 32 - 8.259862 = 23.7

Upper Limit = x + E = 32 + 8.259862 = 40.3

95% CI using t-distribution: 23.7 < µ < 40.3

Hope this helps!

4 0

3 years ago

The fraction ½ can be written as a terminating decimal. true or false?

kvasek [131]

Yes. .50. hope this helps

4 0

3 years ago

Read 2 more answers

Which answer shows 902.6 written in expanded form?

nignag [31]

Since it is 900 + 2 + 0.6
the answer would be A

7 0

4 years ago

Evaluate the following expression: 25x(x - 4) when x = -1