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3 years ago

5

Guided Practice

1 answer:

Nat2105 [25]3 years ago

6 0

Answer:

C.

the number of cans of soda Amy orders

Step-by-step explanation:

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Can you please help me with this question with the steps? <br> 8a^3 -2a^2+ 12a-3

KATRIN_1 [288]

I'm assuming you want to simplify the expression since it's not equal to anything.. you can't solve for "a" unless it's equal to something.

factor

8a³ - 2a²+ 12a - 3

2a²(4a - 1) + 3(4a -1)

(4a - 1) is a like term that can be combined further.

(2a² + 3)(4a - 1)

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4 years ago

What is the volume of a cube with 1/2 inch sides

Studentka2010 [4]

What is the formula for calculating the volume of a cube?

V = s x s x s where s = the length of a side

V = .5 x.5 x.5 = .125

5 0

3 years ago

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Each person in the United States eats about 65 fresh apples a year based on this estimate. How many apples do 3 families of 4 ea

galina1969 [7]

Multiply 3 times 4 which gives you 12 then multiply that by 65 and your answer is 780

8 0

3 years ago

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The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

4 years ago

14 is 70% of what number

emmasim [6.3K]

Move the decimal to the left where 70% becomes (.70). Then divide 14/.70 which equals 20.

8 0

3 years ago

Read 2 more answers