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Romashka [77]
4 years ago
14

Noah wants to create a floor plan that is a scale drawing. The actual length of Wall C is 4 m. To represent Wall C, Noah draws a

segment 16 cm long. What scale is he using?
Mathematics
2 answers:
Viefleur [7K]4 years ago
8 0

Answer:

Step-by-step explanation: it’s 1 m because it’s equal to 4

FinnZ [79.3K]4 years ago
7 0

Answer:

Scale using for plan ⇒ 1 m = 4 cm

Step-by-step explanation:

Given:

Actual length of wall = 4 m

Segment of wall = 16 cm long

Find:

Scale using for plan

Computation:

Actual length of wall = Segment of wall

4 m = 16 cm

1 m = 16 / 4 cm

1 m = 4 cm

Scale using for plan ⇒ 1 m = 4 cm

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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 .

<em>Let </em>\mu<em> = mean self-efficacy score.</em>

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 <u>One-sample z test statistics</u> 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, <u><em>test statistics</em></u>  =  \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%. <u>Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test.</u>

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 <u>we fail to reject our null hypothesis</u>.

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

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3 years ago
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