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nataly862011 [7]

3 years ago

11

salma is 1.45 meters tall. at 3p.m she measures the length of a tree's shadow to 29.35 meters. she stands 24.3 meters away from

the tree, so that the tip of her shadow meets the tip of the treels shadow. find the height of the tree to the nearest hundreth of a meter.

2 answers:

jeka943 years ago

8 0

Answer:

kimjnuhbygvtcfrdxe

Step-by-step explanation:

Sophie [7]3 years ago

7 0

Answer:

23.67

Step-by-step explanation:

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Morton made 36 out of 48 free throws last season. What percent of his free throw did Morton make?

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Just divide 36 by 48, and you will get that he made 75% of his free throws.

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3 years ago

A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Const

Ludmilka [50]

Answer:

$0.609 \leq \sigma \leq 0.772$

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation

s=0.68 represent the sample standard deviation

$\bar x =98.90$ represent the sample mean

n=98 the sample size

Confidence=90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=98-1=97$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:

$\chi^2_{\alpha/2}=120.990$

$\chi^2_{1- \alpha/2}=75.282$

And replacing into the formula for the interval we got:

$\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}$

$0.371 \leq \sigma^2 \leq 0.596$

Now we just take square root on both sides of the interval and we got:

$0.609 \leq \sigma \leq 0.772$

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

3 0

3 years ago

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%

vagabundo [1.1K]

Answer:

1) $\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

2) $p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

$\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}$

Where O rpresent the observed values and E the expected values.

State the null and alternative hypothesis

Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

$E = \% Grand total$

$E_{Catfish}=500*0.3=150$

$E_{Bass}=500*0.15=75$

$E_{Bluegill}=500*0.4=200$

$E_{Pike}=500*0.15=75$

Part 1: Calculate the statistic

$\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

$\chi^2 =16.313$

Calculate the critical value

First we need to calculate the degrees of freedom given by:

$df= (categories-1)=(4-1)= 3$

Since the confidence provided is 95% the significance would be $\alpha=1-0.95=0.05$ and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be $\chi^2_{crit}=7.815$

We can calculate also the p value:

$p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

5 0

2 years ago