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3 years ago

A camp counselor needs 12 bags of candy that each weigh 112 pounds. How many ounces of candy does the camp counselor need?

Mathematics

2 answers:

Svetlanka [38]3 years ago

6 0

I think it is 1,464 ounces he needs

IRISSAK [1]3 years ago

3 0

21,504 ounces hope it helps

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Suppose a different student reaches in the bag, randomly selects their twenty chips, and estimates that 60% of the students are

djyliett [7]

12 chips if you are trying to find 60% of 20

8 0

3 years ago

An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i

Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = $\sqrt{(15-x)^2+1^2}$

The total time, 't', is given as follows;

$t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}$

The minimum value of 't' is found by differentiating with an online tool, as follows;

$\dfrac{dt}{dx} = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} = \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }$

At the maximum/minimum point, we have;

$\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0$

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

$t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89$

Therefore, the distance to run, x ≈ 14.96 mile

6 0

2 years ago

1+12 ddddddddddddddddddddddddddddddddddddddddddddddddddddd

Alex73 [517]

Answer:

Step-by-step explanation:

count on your fingers

6 0

3 years ago

Read 2 more answers

How do I do this please help

Umnica [9.8K]

Answer:

wrong as his calculation was incorrect at initial level itself when he found the result of the division

Step-by-step explanation:

Andre said

3 ÷ $\frac{2}{3}$

$=3 \times \frac{3}{2}$

$=\frac{3 \times 3}{2}$

$=\frac{9}{2}$

but andre calculated it as

$=4\tfrac{1}{3}$

$=\frac{4 \times 3 + 1}{3}$

$=\frac{13}{3}$

Hence his calculation was incorrect at initial level itself

4 0

3 years ago

For a fair coin, suppose you toss the coin 100 times.

Natasha_Volkova [10]

Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

For the binomial distribution, the parameters are given as follows:

n = 100, p = 0.5.

Hence the mean and the standard deviation of the approximation are given as follows:

$\mu = np = 100(0.5) = 50$ .

$\sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5$

Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is <u>one subtracted by the p-value of Z when X = 66.5</u>.

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

$Z = \frac{66.5 - 50}{5}$

Z = 3.3

Z = 3.3 has a p-value of 0.9995.

1 - 0.9995 = 0.0005.

0.0005 = 0.05%

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

6 0

1 year ago