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adell [148]
3 years ago
9

If a container contains 18,000 ounces (oz) of beans, about how many pounds (lb) does it contain? [1 pound = 16 ounces] 70 lb 1,1

25 lb 180 lb 288,000 lb
Mathematics
2 answers:
Margarita [4]3 years ago
8 0

Answer:

The answer is B. 1,125

Step-by-step explanation:


Tom [10]3 years ago
3 0

Answer:

Option b. 1,125 lb

Step-by-step explanation:

A container contains 18,000 ounces ( oz ) of beans.

We have to convert ounces to pounds.

As we know 1 pound = 16 ounces

To convert ounces to pounds we have to divide 18,000 ounces to 16.

18,000 ounces = 18000 ÷ 16

                          = 1,125 lb

Container contains 1,125 pounds of beans.

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x^6-9 = x^-3 = 1/x^2

Step-by-step explanation:

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6 people equally share bill rs 98.10 how ​
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16.35Rs would be paid by everyone.

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98.10/6

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3 years ago
Find the missing side round your answer to the nearest tenth
MissTica

Answer:

x=13.2

Step-by-step explanation:

cos(43)=x/18

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Answered by GAUTHMATH

3 0
3 years ago
Recent survey data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. Their parents must have
astraxan [27]

Answer:

38.76% probability that between 13 and 17 of these young adults lived with their parents

Step-by-step explanation:

I am going to use the normal approxiation to the binomial to solve this question.

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 samples 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)}.

In this problem, we have that:

p = 0.142, n = 125

So

\mu = E(X) = np = 125*0.142 = 17.75

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.142*0.858} = 3.9025

What is the probability that between 13 and 17 of these young adults lived with their parents?

Using continuity correction, this is P(13 - 0.5 \leq X \leq 17 + 0.5) = P(12.5 \leq 17.5), which is the pvalue of Z when X = 17.5 subtracted by the pvalue of Z when X = 12.5. So

X = 17.5

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

Z = \frac{17.5 - 17.75}{3.9025}

Z = -0.06

Z = -0.06 has a pvalue of 0.4761

X = 12.5

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

Z = \frac{12.5 - 17.75}{3.9025}

Z = -1.35

Z = -1.35 has a pvalue of 0.0885

0.4761 - 0.0885 = 0.3876

38.76% probability that between 13 and 17 of these young adults lived with their parents

4 0
3 years ago
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