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svet-max [94.6K]
3 years ago
13

PLEASE HELP AS SOON AS POSSIBLE

Mathematics
1 answer:
Elodia [21]3 years ago
7 0

d = |150 - 20t|

70 = |150 = 20t|

70 = 150 - 20t                -70 = 150 - 20t

-80 = -20t                   -220 = -20t

 4 = t                              11 = t

Answer: 4, 11 seconds

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What is this solution of the equation 4w = 2/3
amid [387]
First, you need to isolate w. So you get w = 2/3 divided by 4.
I made the 2/3 a decimal, as it is easier to work with: 0.6666 divided by 4 is 0.167. 
0.167 is 1/6.
w = 1/6, or 0.167.
6 0
3 years ago
Read 2 more answers
A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea
uysha [10]

Answer:

The 90th percentile for the distribution of the total contributions is $6,342,525.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums of size n, the mean is \mu*n and the standard deviation is s = \sqrt{n}*\sigma

In this question:

n = 2025, \mu = 3125*2025 = 6328125, \sigma = \sqrt{2025}*250 = 11250

The 90th percentile for the distribution of the total contributions

This is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. Then

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

By the Central Limit Theorem

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

1.28 = \frac{X - 6328125}{11250}

X - 6328125 = 1.28*11250

X = 6342525

The 90th percentile for the distribution of the total contributions is $6,342,525.

3 0
3 years ago
How many solutions does this have?​
olga nikolaevna [1]

Answer:

none ok smrbgeriinjhgfd

6 0
3 years ago
Read 2 more answers
Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confide
ioda

Answer:

The question is incomplete, but the step-by-step procedures are given to solve the question.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.005 = 0.995, so Z = 2.575.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.575\frac{\sigma}{\sqrt{n}}

The lower end of the interval is the sample mean subtracted by M.

The upper end of the interval is the sample mean added to M.

The 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (lower end, upper end).

7 0
3 years ago
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Jamal because he has a higher percentage of shots made 78%>62%
4 0
3 years ago
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