the diameter of the sun is 1.4 * 10^-6 m. where as diameter of earth is 1.2756 *10^-7m. compare the diameter of sun and moon

Tyler evaluates the expression -621,044 - (-214,318) and gets an answer of -400,080.

cupoosta [38]

Answer: shut up nerd you wont report me you goodie toshoos

Step-by-step explanation:

7 0

4 years ago

What is 7.25 / 1,000

Serggg [28]

Answer:.00725

Step-by-step explanation:

8 0

3 years ago

Can some one help me solve this ?

kondaur [170]

<h2>The answer is d.</h2>

v = lwh

v/lh = lwh/lh

w = v/lh

5 0

3 years ago

Determine the slope of the line that contains the points E(6,3) & F(-6,3).

vitfil [10]

Count the spaces between the points rise over run

6 0

3 years ago

A clinical psychologist wants to test whether experiencing childhood trauma reduces one's self-efficacy in adulthood. He randoml

lidiya [134]

Answer:

No, there is not any sufficient evidence to conclude that individuals who have experienced childhood trauma have lower self-efficacy in adulthood.

Step-by-step explanation:

We are given that a clinical psychologist wants to test whether experiencing childhood trauma reduces one's self-efficacy in adulthood.

He randomly selects 22 adults who have experienced childhood trauma and finds that their mean self-efficacy score equals 118.1.

Self-efficacy scores in the general population of adults are distributed normally with a mean equal to 118.5 and a standard deviation equal to 18.8 .

Let $\mu$ = mean self-efficacy score.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 118.5 {means that the individuals who have experienced childhood trauma have higher or same self-efficacy in adulthood}

Alternate Hypothesis, $H_A$ : $\mu$ < 118.5 {means that the individuals who have experienced childhood trauma have lower self-efficacy in adulthood}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean self-efficacy score = 118.1

$\sigma$ = population standard deviation = 18.8

n = sample of adults who have experienced childhood trauma = 22

So, test statistics = $\frac{118.1-118.5}{\frac{18.8}{\sqrt{22} } }$

= -0.0998

The value of z test statistics is -0.0998.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.0998 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the individuals who have experienced childhood trauma have higher or same self-efficacy in adulthood.

4 0

3 years ago