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maxonik [38]
3 years ago
14

Lily has 3500 text per month on her cell phone plan up to how many more task can she spend if she already spent 2612 text

Mathematics
2 answers:
bazaltina [42]3 years ago
8 0

Answer:

888 more Text

Step-by-step explanation:

maria [59]3 years ago
4 0

Answer: She has up to 888 texts left

Step-by-step explanation:

3500-2612=888

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What is the probability of tossing a coin and getting a head 4 times in a row?
alexira [117]

Answer:

D

Step-by-step explanation:

6 0
3 years ago
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Which of the following is an example of empirical probability
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Just wondering, Is there options?
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Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1
svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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 normal 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.

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.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

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

By the Central Limit Theorem

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

Z = \frac{9.3 - 8}{0.4919}

Z = 2.64

Z = 2.64 has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

n = 6 means that s = \frac{1.1}{\sqrt{6}} = 0.4491

This probability is 1 subtracted by the pvalue of Z when X = 9. So

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

Z = \frac{9 - 8}{0.4491}

Z = 2.23

Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

8 0
3 years ago
Evaluate the function f(x) at the given numbers (correct to six decimal places).
statuscvo [17]

Answer:

The values of given function are shown in the below table.

Step-by-step explanation:

The given function is

f(x)=\frac{x^2-5x}{x^2-x-20}

Simplify the given function.

f(x)=\frac{x(x-5)}{x^2-5x+4x-20}

f(x)=\frac{x(x-5)}{x(x-5)+4(x-5)}

f(x)=\frac{x(x-5)}{(x+4)(x-5)}

Cancel out the common factor.

f(x)=\frac{x}{x+4}

Substitute x=5.5 in the above equation.

f(5.5)=\frac{5.5}{5.5+4}

f(5.5)=\frac{5.5}{9.5}

f(5.5)=0.57894736842

f(5.5)\approx 0.578947

Similarly find the value for all values of x.

The values of given function are shown in the below table.

8 0
3 years ago
40lb is what percent of 75 lb
andreev551 [17]

Answer:

i got 53.3%

Step-by-step explanation:

The formula is

is over of = what the percent is over 100

4 0
3 years ago
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