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2 years ago

13

Little helppp please

1 answer:

liberstina [14]2 years ago

8 0

Answer:

Answer given above

Step-by-step explanation:

a.) 6 m n^8

b.)2x^8 y^3

c. ) d^3 e

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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

3 years ago

Simplify 8(xy + 8) + 1. 8xy + 9 8xy + 65 8xy + 73

Advocard [28]

Answer:

8xy + 65

Step-by-step explanation:

8(xy + 8) + 1.

8 x 8=

64 + 1=

65 (8xy + 65)

---------------------------------------------

LINE THE NUMBERS UP AND IT WILL MAKE SENSE

5 0

3 years ago

A professor grades students on four tests, a term paper, and a final examination. Each test counts as 15% of the course grade. T

Dmitry [639]

Answer:

Marilee wants to earn an "A" in a class and needs an overall average of at least Her test grades are 88 , 92,100 , and 80 . The average of her quizzes is 90 and counts as one test grade. The final exam counts as 2.5 test grades. What scores on the final exam would result in Marilee's overall average of 92 or greater? (See Example 9

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3 years ago

24, 43, 89, 15 Which two numbers are prime numbers?

Nikitich [7]

Answer: 43 and 49

Step-by-step explanation:

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3 years ago

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Find the x and y intercept. 3x + 2y = 12

iris [78.8K]

Answer:

x=4 y=6

Step-by-step explanation:

12 divide by 3 = 4

12 divide by 2 = 6

4 0

3 years ago

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