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3 years ago

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Solve this problem in your notebook using all four steps. Harvey is 3 times as old as Jane. The sum of their ages is 48 years. F

ind the age of each. Jane is a0 years old. Harvey is a1 years old.

1 answer:

JulsSmile [24]3 years ago

3 0

Harvey is 3 times as old as Jane."
<span>h = 3j </span>

<span>"The sum of their ages is 48 years." </span>
<span>h + j = 48 </span>

<span>substitution </span>
<span>(3j) + j = 48 </span>
<span>4j = 48 </span>
<span>j = 12 </span>
<span>h = 3j = 36</span>

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Answer:

a. 341.902.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

n instances of a normal variable:

For n instances of a normal variable, the mean is $n\mu$ and the standard deviation is $s = \sigma\sqrt{n}$

60 days, for each day, mean 6, variance of 12.

So

$\mu = 60*6 = 360$

$s = \sqrt{12}\sqrt{60} = 26.8328$

What is the 25th percentile of her total wait time over the course of 60 days?

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 360}{26.8328}$

$X - 360 = -0.675*26.8328$

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Thus, the correct answer is given by option A.

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3 years ago

In finance the notion of expected value is used to analyze investments for which the investor has an estimate of the chances ass

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Answer:

Step-by-step explanation:

The expected return is given as

Expected Return = SUM (Return i x Probability i). i=1,2,3.....

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Probably of 0.3, it loses 20cents per dollar.

Expected return=(0.7×60)-(0.3×20)

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To dollars, 1cents is 0.01dollars

Then, 36cents = 0.36dollars

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3 years ago

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

8 0

3 years ago

Gala2k [10]

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

<u>Given formula:</u>

$\sf N=N_0e^{-kt}$

$\sf N_0$ = initial mass (at time t=0)
N = mass (at time t)
k = a positive constant
t = time (in days)

Given values:

$\sf N_0$ = 11 g
k = 0.125

Half-life: The <u>time</u> required for a quantity to reduce to <u>half of its initial value</u>.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

brainly.com/question/28016999

5 0

2 years ago

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