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1 year ago

8

If the hypotenuse of a right triangle is 2

lt="\sqrt{3}" align="absmiddle" class="latex-formula"> units long and one of the legs is $\sqrt{3}$ units long, then how long is the other leg?
$\sqrt{3}$

2 $\sqrt{3}$

3

1 answer:

Svetradugi [14.3K]1 year ago

5 0

Answer:

3

Step-by-step explanation:

a = the other leg:

$a^{2} =(2\sqrt{3} )^{2} -(\sqrt{3}) ^{2} =(4)(3)-3=12-3$

$a^{2} =9$

$a=\sqrt{9}$

$a=3$

Hope this helps

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It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele

kow [346]

Answer:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

Step-by-step explanation:

1) Data given and notation

$\bar X=23500$ represent the sample mean

$s=3900$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =2000$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2000$

Alternative hypothesis: $\mu > 2000$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

Calculate the P-value

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

$df=n-1=100-1=99$

Since is a one-side upper test the p value would be:

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

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3 years ago

A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect

krek1111 [17]

Answer:

The probability that the first two electric toothbrushes sold are defective is 0.016.

Step-by-step explanation:

The probability of an event, say <em>E </em>occurring is:

$P (E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let <em>X</em> = number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is, <em>n</em> = 20.

Number of defective electric toothbrushes is, <em>x</em> = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

${20\choose 2}=\frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}=\frac{20\times 19\times 18!}{2!\times 18!}=190$

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

${3\choose 2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\times 1!}=3$

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

$=\frac{3}{190}\\$

$=0.01579\\\approx0.016$

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

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Answer:

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Answer:

Step-by-step explanation:

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