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Elodia [21]
3 years ago
15

Write the percent as a decimal 152%

Mathematics
2 answers:
Tpy6a [65]3 years ago
7 0

To write a percent as a decimal, first remember that a percent is ratio that compares a number to 100. We can think of 152% as the ratio 152 to 100 or \frac{152}{100}. Remember that dividing by 100 moves the decimal point 2 places to the left.

schepotkina [342]3 years ago
4 0

1.52 would be the answer. Hope i helped.


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AVprozaik [17]

Answer:

x = 65, y = 63

Step-by-step explanation:

∠ EGA = ∠ HBG = 52° ( corresponding angles )

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yulyashka [42]

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4 0
3 years ago
Use the given information to find the minimum sample size required to estimate an unknown population mean mu. How many students
babymother [125]

Answer:

9413

Step-by-step explanation:

Given that :

Error = $2 ; Standard deviation (σ) =99 ; α = 95% = 0.95

Zα/2 = Z0. 05/2 = 1.96 (Z probability calculator)...

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n = ((Zα/2 * σ) / Error)^2

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5 0
3 years ago
3. Suppose the local Best Buy store averages 301 customers every day entering the facility with a standard deviation of 80 custo
stich3 [128]

Answer:

The probability that the average number of customers in the sample is between 290 and 310 is 0.61922.

Step-by-step explanation:

We are given that the local Best Buy store averages 301 customers every day entering the facility with a standard deviation of 80 customers.

A random sample of 50 business days was selected.

<em>Let </em>\bar X<em> = sample average number of customers</em>

The z-score probability distribution for sample average is given by;

             Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population average = 301 customers

            \sigma = population standard deviation = 80 customers

            n = sample of business days = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

(a) Probability the average number of customers in the sample is between 290 and 310 is given by = P(290 < \bar X < 310) = P(\bar X < 310) - P(\bar X \leq 290)

   P(\bar X < 310) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{310-301}{\frac{80}{\sqrt{50} } } ) = P(Z < 0.79) = 0.78524

   P(\bar X \leq 290) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{290-301}{\frac{80}{\sqrt{50} } } ) = P(Z \leq -0.97) = 1 - P(Z < 0.97)

                                                       = 1 - 0.83398 = 0.16602                       

<em>{Now, in the z table the P(Z  </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.79 and x = 0.97 in the z table which has an area of 0.78524 and 0.83398 respectively.}</em>

Therefore, P(290 < \bar X < 310) = 0.78524 - 0.16602 = 0.61922

<u><em>Hence, the probability that the average number of customers in the sample is between 290 and 310 is 0.61922.</em></u>

4 0
3 years ago
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