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Pachacha [2.7K]
3 years ago
14

HELPPP !!! ILL MARL BRAINLIST

Mathematics
1 answer:
almond37 [142]3 years ago
3 0

Answer:

Hi there!

Your answers are:

1) 2^4 = 2×2×2×2 = 16

2) 2^3= 2×2×2= 8

3) 2^2 = 2×2 = 4

4) 2^1 = 2

5)2^0 = 1

6)2^ -1 = .5

7) 2^ -2 = .25

8) 2^ -3 = .125

9) 2^ -4= .0625

Hope this helps

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4x-7x=13 in a fraction
Naddika [18.5K]
4x-7x= 13
⇒ -3x= 13
⇒ x= 13/(-3)
⇒ x= -13/3

The final answer is x= -13/3~
4 0
3 years ago
(64-8)÷4 answer ASAP please​
Ann [662]

Answer:

14

Step-by-step explanation:

64-8=56

56/4=14

4 0
3 years ago
The time taken to assemble a laptop computer in a certain plant is a random variable having a normal distribution of 20 hours an
ludmilkaskok [199]

Answer:

a) 40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b) 34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

\mu = 20, \sigma = 2

a)Less than 19.5 hours?

This is the pvalue of Z when X = 19.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{19.5 - 20}{2}

Z = -0.25

Z = -0.25 has a pvalue of 0.4013.

40.13% probability that a laptop computer can be assembled at this plant in a period of time of less than 19.5 hours.

b)Between 20 hours and 22 hours?

This is the pvalue of Z when X = 22 subtracted by the pvalue of Z when X = 20. So

X = 22

Z = \frac{X - \mu}{\sigma}

Z = \frac{22 - 20}{2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 20

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 20}{2}

Z = 0

Z = 0 has a pvalue of 0.5

0.8413 - 0.5 = 0.3413

34.13% probability that a laptop computer can be assembled at this plant in a period of time between 20 hours and 22 hours.

4 0
3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

7 0
3 years ago
Which is equivalent to 64 1/4
professor190 [17]
It A i just took the quiz and got it but fail 

7 0
3 years ago
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