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1 year ago

7

The mean life of a television set is 119 months with a standard deviation of 13 months. If a sample of 67 televisions is randoml

y selected, what is the probability that the sample mean would differ from the true mean by less than 2 months? Round your answer to four decimal places.

1 answer:

PIT_PIT [208]1 year ago

5 0

Answer:

Provided in the picture below.

Step-by-step explanation:

Provided in the picture below.

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Two race cars,car x an y,are at the starting point of a two km track at the same time.car x and car y make one lap every 60 s an

tino4ka555 [31]

Answer is attached.

Please check the image.

4 0

2 years ago

A school cafeteria sells milk at 25 cents per carton and salads at 45 cents each. one week the total sales for these items were

denis-greek [22]

solution:

Lets start with the most amount that could have been sold.......using guess and check, we can figure out that 290 salads could have been sold, while 8 cartons of milk would have been sold.

The least amount of salads that could have been sold were none.

so,

you have 0<s<290

at least none were sold, and at most 290 were sold

but I do believe you are missing part of the question

4 0

2 years ago

Find the area of the part of the plane 5x + 4y + z = 20 that lies in the first octant.

BlackZzzverrR [31]

This part of the plane is a triangle. Call it $\mathcal S$ . We can find the intercepts by setting two variables to 0 simultaneously; we'd find, for instance, that $y=z=0$ means $5x=20\implies x=4$ , so that (4, 0, 0) is one vertex of the triangle. Similarly, we'd find that (0, 5, 0) and (0, 0, 20) are the other two vertices.

Next, we can parameterize the surface by

$\mathbf s(u,v)=\langle4(1-u)(1-v),5u(1-v),20v\rangle$

so that the surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=20\sqrt{42}(1-v)\,\mathrm du\,\mathrm dv$

Then the area of $\mathcal S$ is given by the surface integral

$\displaystyle\iint_{\mathcal S}\mathrm dS=20\sqrt{42}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-v)\,\mathrm dv\,\mathrm du$
$\displaystyle=20\sqrt{42}\int_{v=0}^{v=1}(1-v)\,\mathrm dv=10\sqrt{42}\approx64.8074$

3 0

2 years ago

The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.

Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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.

If $Z \leq -2$ or $Z \geq 2$ , the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1$

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

$Z = \frac{19 - 22}{1}$

$Z = -3$

$Z = -3 \leq -2$ , so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0

1 year ago

Which expression can be used to find the sum of the polynomials? (9 – 3x2) (–8x2 4x 5)

SIZIF [17.4K]

For this case we have:

Polynomial 1: $P (x) = 9-3x ^ 2$

Polynomial 2: $Q (x) = - 8x ^ 2 + 4x + 5$

Sorting the polynomials:

Polynomial 1: $P (x) = - 3x ^ 2 + 9$

Polynomial 2: $Q (x) = - 8x ^ 2 + 4x + 5$

Adding term to term (similar) we have:

$P (x) + Q (x) = (- 3-8) x ^ 2 + (0 + 4) x + (9 + 5)\\P (x) + Q (x) = - 11x ^ 2 + 4x + 14$

Answer:

$A (x) = P (x) + Q (x) = - 11x ^ 2 + 4x + 14$

7 0

1 year ago

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