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2 years ago

Please help beautifuls!

Mathematics

1 answer:

jok3333 [9.3K]2 years ago

5 0

Answer:

d-2

Step-by-step explanation:

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HELP ME!!!! ASAP!!! PLEASE!!!<br> Explain why the equation (x-6)^2-3=13 has two solutions.

Yuki888 [10]

It has 2 equations bc it hasq 2 part

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3 years ago

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Find b: If a = 9 and c = 15<br> С<br> a<br> b

pogonyaev

Answer:

b=12

Step-by-step explanation:

a^2+b^2=c^2

(9×9) + b^2= 15×15

81+ b^2= 225

-81. -81

b^2=144

√. √

b= 12

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3 years ago

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A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and

gayaneshka [121]

Answer:

$(832.156, \ 847.844)$

Step-by-step explanation:

Given data :

Sample standard deviation, s = 15

Sample mean, $\overline x = 840$

n = 23

a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}$

$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$

$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$

∴ $E = 2.508 \times \frac{15}{\sqrt{23}}$

$E = 7.844$

So, 98% CI is

$(\overline x - E, \overline x + E)$

$(840-7.844 , \ 840+7.844)$

$(832.156, \ 847.844)$

4 0

3 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

3 years ago

last year the depth of the river was 4.2 feet deep. this year it dropped 24% find the depth of the river this year

zzz [600]

Answer:

3.2

Step-by-step explanation:

7 0

2 years ago