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Usimov [2.4K]
2 years ago
15

Show Your Work! Will Mark Brainliest!

Mathematics
1 answer:
Nikolay [14]2 years ago
6 0

Answer:

x= -4

Step-by-step explanation:

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Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2
r-ruslan [8.4K]

Answer:

The value of given expression is \frac{21}{128}.

Step-by-step explanation:

Given information: n=7, x=2, p=1/2

q=1-p=1-\frac{1}{2}=\frac{1}{2}

The given expression is

C(n,x)p^xq^{n-x}

It can be written as

^nC_xp^xq^{n-x}

Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.

^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}

\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}

\frac{7!}{2!5!}(\frac{1}{2})^{2+5}

\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}

21(\frac{1}{2})^{7}

\frac{21}{128}

Therefore the value of given expression is \frac{21}{128}.

7 0
3 years ago
80 81 82 83 84 85 86 87 88 89 90
cluponka [151]

Answer:

Step-by-step explanation:

Um... there is no "Lorenzo's scores."

8 0
3 years ago
The scale drawing represents Kim’s living room. She wants to put new baseboards on the walls of the room. If the scale is 2 cm:1
Serga [27]
The answer is B, because if you draw the baseboards on the scale drawing, you'll get nineteen baseboards. 19 x 4,19 = 79,61

5 0
3 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}}  } ~N(0;1)

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
3. The amount was $120 at 5% rate for 4<br> years.
Inga [223]

Answer:

Go to photo math thank me later

Step-by-step explanation:

dowload tha app on ur phone

5 0
3 years ago
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