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OLga [1]
3 years ago
5

Match wash hypotenuse length with the leg lengths that will create a right triangle

Mathematics
2 answers:
Ugo [173]3 years ago
5 0
Hope this can help you

Stells [14]3 years ago
3 0

Answer:

Step-by-step explanation:

Using the Pythagoras theorem, we have

(Hyp)^2=(Leg)^2+(Leg)^2

(A) The given value of hypotenuse is:

\sqrt{6} units

Now, using the Pythagoras theorem,

\sqrt{6}=\sqrt{(\sqrt{5})^2 +(1)^2}

Thus, one leg is \sqrt{5} units and the other is 1 units.

(B)The given value of hypotenuse is:

\sqrt{5} units

Now, using the Pythagoras theorem,

\sqrt{5}=\sqrt{(\sqrt{2})^2 +(\sqrt{3})^2}

Thus, one leg is \sqrt{2} units and the other is \sqrt{3} units.

(C) The given value of hypotenuse is:

\sqrt{8} units

Now, using the Pythagoras theorem,

\sqrt{8}=\sqrt{(\sqrt{5})^2 +(\sqrt{3})^2}

Thus, one leg is \sqrt{5} units and the other is \sqrt{3} units.

(D) The given value of hypotenuse is:

\sqrt{3} units

Now, using the Pythagoras theorem,

\sqrt{3}=\sqrt{(\sqrt{2})^2 +(1)^2}

Thus, one leg is \sqrt{2} units and the other is 1 units.

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Uma is paid $12.25 for each chair she decorates. Last week she
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$514.50

Step-by-step explanation:

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6 0
3 years ago
Suppose that in a random selection of 100 colored​ candies, 28​% of them are blue. The candy company claims that the percentage
Zigmanuir [339]

Answer:

We conclude that the percentage of blue candies is equal to 29​%.

Step-by-step explanation:

We are given that in a random selection of 100 colored​ candies, 28​% of them are blue. The candy company claims that the percentage of blue candies is equal to 29​%.

Let p = <u><em>population percentage of blue candies</em></u>

So, Null Hypothesis, H_0 : p = 29%     {means that the percentage of blue candies is equal to 29​%}

Alternate Hypothesis, H_A : p \neq 29%     {means that the percentage of blue candies is different from 29​%}

The test statistics that will be used here is <u>One-sample z-test for</u> <u>proportions</u>;

                         T.S.  =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of blue coloured candies = 28%

           n = sample of colored​ candies = 100

So, <u><em>the test statistics</em></u> =  \frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }

                                    =  -0.22

The value of the z-test statistics is -0.22.

<u>Also, the P-value of the test statistics is given by;</u>

               P-value = P(Z < -0.22) = 1 - P(Z \leq 0.22)

                            = 1 - 0.5871 = 0.4129

Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29​%.

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3 years ago
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3 years ago
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Katen [24]

Answer:

9

Step-by-step explanation:

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2 years ago
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