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3 years ago

8

Mathematics

1 answer:

STatiana [176]3 years ago

5 0

Answer:

(e^3 / e^2n)

Step-by-step explanation:

isn't this the same question

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19. Suppose 3 marbles are drawn without replacement from a bag containing 3 yellow and 4 white marbles. (a) Draw a histogram for

Alina [70]

Answer:

(a) see attachment

(b) E(x) = 226/105

Step-by-step explanation: see attachment

7 0

3 years ago

What is the value of x in the number 5/3+ 3x+4=8

amm1812

5/3+3x+4= 8
= 3x= 4-5/3
= 3x= 7/3
= x= 7/9

4 0

2 years ago

Haley practiced her free throws at the basket ball court and shot 25 times. She made 11 of her shots. What percent of her shots

ivolga24 [154]

Answer:

Step-by-step explanation:

Haley practiced her free throws at the basket ball court and shot 25 times

She made 11 of her shots.

find: What percent of her shots did she NOT make?

<u>11 shots made</u> = 44%

25 shots

100% - 44% = 56%

therefore, 56% of her shot did NOT make it.

8 0

2 years ago

Wanda wants to buy a pair of shoes that cost $48 if sales tax is 8.25% what will be the total cost

lidiya [134]

48 x 8.25% = 3.96
Now add 3.96 + 48
Answer 51.96

3 0

3 years ago

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

$\= x = 13.4$ . $E = 2.3$

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

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

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

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

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

Generally the margin of error is mathematically represented as

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

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

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

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

3 0

3 years ago