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3 years ago

What is 1 over 200 as a percent

1 answer:

6 0

The answer is 0.5

A percentage is over 100.

So rewrite 1/200 with a denominator of 100, then the numerator is the percent.

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The candle factory made 498 red candles and 204 white candles each day. About how many candles did the factory make in a day alt

stiv31 [10]

Answer:

about 700

Step-by-step explanation:

you add 204 + 498=

4 0

2 years ago

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Help me solve this please ill give good rating...

blagie [28]

Answer: For the first part, her total savings would be $88, and for the second part, (let's pretend s = the total savings) the equation would be s = 40 + w x 6.

Step-by-step explanation: As for the first part, we would need to multiply her money per week times the amount of weeks she works. We can do this by simply multiplying the amount she makes (6) by the amount of weeks she works (8), resulting in 48, but we still have to add that number to the amount she already has, or 40, making $88 in total, as for the second part, this does a great job of explaining the reasoning behind that as well. Hope this answered your question!

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3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

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2 years ago

Which best describes the relationship between the successive terms in the sequence below? –3.2, 4.8, –7.2, 10.8, …

Leto [7]

Since you are not given the following choices, by analyzing this series, it is a geometric progression because their common ratio is constant. Proofing that
4.8/(-3.2) = -1.5, also
-7.2/(4.8) = -1.5, and also
10.8/(-7.2) = -1.5

6 0

2 years ago

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Solve the equation 3x - 2(x + 1) = 2x - 7

natta225 [31]

Answer:

X=5

Step-by-step explanation:

Collect like terms then move the terms. Then collect liked terms. Calculate and Chang signs

3 0

2 years ago

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