Multiples of six and 8

Mathematics

1 answer:

finlep [7]3 years ago

4 0

Answer:

Multiples of 6: 6, 12, 18, 24, 30, 36 Multiples of eight:8, 16, 24, 32, 40,

Step-by-step explanation:

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3 years ago

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bekas [8.4K]

Answer:

Question 6a: $p \leq 16$

Question 6b: option d

Step-by-step explanation:

Question 6a: $6p \leq 96\\6p/6 \leq 96/6\\p \leq 16$

Question 6b:

- closed dot since its equal to 16

- p is less than 16 (arrow pointing left)

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3 years ago

The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$