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3 years ago

PLZ HELP! WILL GIVE BRAINLIEST TEN POINTS

Mathematics

2 answers:

dexar [7]3 years ago

8 0

Answer:

48,200 hours

Tatiana [17]3 years ago

3 0

Answer:

Time = distance/speed. T = 1,200,000,000/28,000. This can be cancelled to 1,200,000/28. Rounding 28 to 30 for the estimate, divide by another ten: 120,000. Then divide by the remaining 3 from 30: 120,000/3 = 40,000. It will take approximately 40,000 hours.

<em>Hope This Helps!</em>

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How many ounces of pure gold and a 16 carry the old change and it weighs 2.5 ounces

sladkih [1.3K]

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3 years ago

Find the number of letters from the word MAXIMUM where two consonants cannot occur. (Permutation)

Hunter-Best [27]

Answer:

Step-by-step explanation:

In word MAXIMUM, consonants are M & X. According to the question itself, consonants should not occur together.

So, we have to put vowels at each even position; only then the condition will be satisfied.

We can put this in 3! ways (permutations of A, I & U).

For odd positions we can put 3-M's & a X. Total ways is 4!/3! (as repetitions of M).

Hence, the answer is 4!/3! x 3! = 24 ways.

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2 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$