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3 years ago

What % of 19.6 is 10.78?

1 answer:

5 0

19.6=100%

Therefore:

19.6----------------------100%
10.78-------------------- x

x=(10.78 * 100%) / 19.6=55%

Answer: 55% of 19.6 is 10.78

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Standard form for y=mx+b

butalik [34]

Answer:

that's standard form for slope

Step-by-step explanation:

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3 years ago

Whats the answer to this problem 9x+10=2001-X

Vilka [71]

Answer: x=1991/10

Step-by-step explanation: Step 1: Simplify both sides of the equation.

9x+10=2001−x

9x+10=2001+−x

9x+10=−x+2001

Step 2: Add x to both sides.

9x+10+x=−x+2001+x

10x+10=2001

Step 3: Subtract 10 from both sides.

10x+10−10=2001−10

10x=1991

Step 4: Divide both sides by 10.

10x

1991

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3 years ago

Read 2 more answers

A student estimated a mass to be 250g but upon carefully measuring it found the actual mass is 240g what is the percentage error

igomit [66]

Maybe 4.167% I'm not sure sorry

6 0

3 years ago

Find the amount necessary to fund the given withdrawals.

garri49 [273]

Answer:

the amount that necessary to fund is $2,671.61

Step-by-step explanation:

The computation of the amount is shown below;

Given that

PMT = $270

NPER = 6 × 2 = 12

RATE = 6.2% ÷ 2 = 3.1%

FV = $0

The formula is shown below:

=-PV(RATE;NPER;PMT;FV;TYPE)

After applying the above formula the present value is $2,671.61

hence, the amount that necessary to fund is $2,671.61

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3 years ago

Consider the probability that no less than 95 out of 152 registered voters will vote in the presidential election. Assume the pr

nikdorinn [45]

Answer:

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given registered voter will vote in the presidential election is 61%.

This means that $p = 0.61$

152 registed voters:

This means that $n = 152$

Mean and Standard deviation:

$\mu = E(X) = 152*0.61 = 92.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{152*0.61*0.39} = 6.01$

Probability that no less than 95 out of 152 registered voters will vote in the presidential election.

This is, using continuity correction, $P(X \geq 95 - 0.5) = P(X \geq 94.5)$ , which is 1 subtracted by the pvalue of Z when X = 94.5. So

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

$Z = \frac{94.5 - 92.72}{6.01}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

1 - 0.6179 = 0.3821

0.3821 = 38.21% probability that no less than 95 out of 152 registered voters will vote in the presidential election.

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3 years ago