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3 years ago

Solve for x. Round to the nearest tenth, if necessary.

Mathematics

2 answers:

Yakvenalex [24]3 years ago

8 0

X=3.8

Sin41/2.5 = sin90/x
Sin90 x 2.5 = sin41 x X
Sin90 x 2.5/ sin41
X=3.8

Igoryamba3 years ago

5 0

Answer:

3.8106327168

Step-by-step explanation:

x=2.5/sin(41) = 3.81063271676

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1 year ago

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$

= $P(- 0.7< Z< 2.1)$

= $(Z< 2.1) - (Z< -0.7)$

= 0.9821 - .2420

= 0.7401

3 0

3 years ago

THIS IS FUNNY

yaroslaw [1]

Answer:

thats funny lol :)

Step-by-step explanation:

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2 years ago