State if each pair of ratios forms a proportion 4/3 and 8/6 yes or no??

On a trip, a car traveled 80 miles in an hour and a half, then was stopped in traffic for 30 minutes, then traveled 100 miles du

Wewaii [24]

Answer:

45 mph

Step-by-step explanation:

6 0

3 years ago

Is 11^10 equivalent to 11^5

oee [108]

Answer:

No

Step-by-step explanation:

They are different because the base are the same but the exponents are different.

6 0

3 years ago

The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh

MAXImum [283]

Answer:

a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`= 6,700 hours

Sample standard deviation=s= 600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282

Z(0.05)=1.645

Z(0.025)=1.960

t(0.01)(49)= 1.299

t(0.05)= 1.677

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance, there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0 for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test we reject H0 as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ± z∝/2 (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0

3 years ago

What is the ratio of sharpened pencils to unsharpened pencils

KatRina [158]

If 20 pencils are sharpened, and there was 36 to begin with, the answer would be 4:5

8 0

3 years ago

Oliver interviewed 30% of the 9th grade class and 70% of the 10th grade class at his school. Jenny interviewed 75% of the 9th gr

mixer [17]

Answer:

A . 36

Step-by-step explanation:

We are given a total of 176 interviewed by Oliver and a total of 140 interviewed by Jenny. To find how many more 10th graders than 9th graders were interviewed, subtract the totals given

176 - 140 = 36

This is how we came to the answer:

We are given 70% of the 10th-grade and 30% of the 9th-grade with a total of 176 for Oliver.

While we're given 75% of the 9th-grade class and 25% of the 10th-grade with a total of 140 interviewed by Jenny

<u>Oliver's Interviewees</u>

10-graders

Firstly, let's find what the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

70% of 176 =

$\frac{70}{100} * \frac{176}{1}$

Cross multiply

123.2 were 10-graders interviewed by Oliver

9th-graders

Now, to find the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

30% of 176 =

$\frac{30}{100} * \frac{176}{1}$

Cross multiply

52.8 were 9th-graders interviewed by Oliver

<u>Jenny's Interviewees</u>

9th-graders

Firstly, let's find what the number of 9th-graders was interviewed by Jenney; find the percentage of the 9th-graders by the total;

75% of 140 =

$\frac{75}{100} * \frac{140}{1}$

Cross multiply

105 students were 9th-graders interviewed by Jenney.

10th-graders

Now, to find the number of 10th-graders was interviewed by Jenney; find the percentage of the 10th-graders by the total;

25% of 140 =

$\frac{25}{100} * \frac{140}{1}$

Cross multiply

35 students were 10th-graders interviewed by Jenney.

<u />

<u>Total calculation</u>

Use the results and sum them up by 9th-grade plus 9th-grade and 10th-grade plus 10-grade. Then subtract the amount gotten from 9th-grade away from the amount gotten from 10th-grade;

Oliver's 9th-grade = 52.8

Jenny's 9th-grade = 105

105 + 52.8 = 157.8

Oliver's 10th-grade = 123.2

Jenny's 10th-grade = 35

123.2 + 35 = 158.2

Total calculation: 158. 2 - 157.8 = 0.4

<h2>Therefore, there are 36 more 10th than 9th.</h2>

<u />

<h3><u>Extra Info:</u></h3>

<u>Oliver's Interviewees Percentage</u>

Since we are given 30% of the 9th-grade class and 70% of the 10th-grade class, first, let's add the percentages. To do so, set it up as a fraction;

30% = $\frac{30}{100}$ while, 70% = $\frac{70}{100}$

Now solve it;

$\frac{30}{100} + \frac{70}{100}$

Simplify; cancel bottom zero's;

$\frac{30}{1} + \frac{70}{1}$

Add the remaining numerators;

30 + 70 = 100

Which is 100%

<u>Jenny's Interviewees Percentage</u>

Since we're given 75% of the 9th-grade class and 25% of the 10th-grade, it will end up the same answer. I'll show you how; first, let's add the percentages. To do so, set it up as a fraction;

25% = $\frac{25}{100}$ and, 75% = $\frac{75}{100}$

Now solve it;

$\frac{25}{100} + \frac{75}{100}$

Simplify; cancel bottom zero's

$\frac{25}{1} + \frac{75}{1}$

Add the remaining numerators;

25 + 75 = 100

Meaning 100%

5 0

3 years ago