Which expression is equivalent to (2x4y)3 2x7y4 2x12y3 8x7y4 8x12y3

What is the expanded 97,367

Ivahew [28]

Answer:

Expanded Notation Form:

97,367 =

90,000

+ 7,000

+ 300

+ 60

+ 7

Expanded Factors Form:

97,367 =

9 × 10,000

+ 7 × 1,000

+ 3 × 100

+ 6 × 10

+ 7 × 1

Expanded Exponential Form:

97,367 =

9 × 104

+ 7 × 103

+ 3 × 102

+ 6 × 101

+ 7 × 100

3 0

3 years ago

Read 2 more answers

Please help me i'm being timed

Evgen [1.6K]

To Find :-

The algebraic equations to solve the above problem .

Solution :-

Let ,

Large number = L
Small number = S

— According to first statement ,

L - S = 4
L - S - 4 = 0

— According to second statement ,

3L + 4S = 54
3L + 4S - 54 = 0 .

These equations matches with the second option b .

Option b is the correct answer.

7 0

3 years ago

Wonder if someone can help me with my math work please

stellarik [79]

Answer:

The slope is negative. The relationship is proportional

Step-by-step explanation:

6 0

3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .