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Diano4ka-milaya [45]
3 years ago
11

A television network is about to telecast a new television show. Before a show is launched, the network airs a pilot episode and

receives a report assessing favorable or unfavorable viewer response. In the past, 60% of the network's shows have received a favorable response from viewers, and 40% have received an unfavorable response. If 50% of the network’s shows have received a favorable response and have been successful, and 30% of the network’s shows have received an unfavorable response and have been successful, what is the probability that this new show will be successful if it receives a favorable response? A. 0.41 B. 0.53 C. 0.67 D. 0.70 E. 0.83
Mathematics
1 answer:
GalinKa [24]3 years ago
5 0

Let's first denote:

F - a favorable response

F' - an unfavorable response

S - successful

We know that:

P(F) = 0.6\\\\P(F')=0.4\\\\P(F\cap S)=0.5\\\\P(F'\cap S)=0.3

So, from the conditional probability, we can calculate:

P(S|F) = \dfrac{P(S\cap F)}{P(F)} = \dfrac{0.5}{0.6}=\dfrac{5}{6}\approx\boxed{0.83}

Answer E.

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Answer:

  No

Step-by-step explanation:

A probability is a unitless value between 0 and 1, inclusive. 9.48 cannot be a probability (too big). 0.4 acres cannot be a probability (has units).

4 0
3 years ago
A car dealership sold 84 cars in April. The dealership wants to increase the number of cars sold by 25% in May. How many cars wi
Pie

Answer: 105 Cars

Step-by-step explanation: To get an increase of 25% you need to multiply 84 by .25 then add the product of that to 84 to get 105

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2 years ago
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A sample of 5 buttons is randomly selected and the following diameters are measured in inches. Give a point estimate for the pop
Helen [10]

Answer:

s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

But we need to calculate the mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_I}{n}

And replacing we got:

\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072

And for the sample variance we have:

s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance \sigma^2

E(s^2) = \sigma^2

Step-by-step explanation:

For this case we have the following data:

1.04,1.00,1.13,1.08,1.11

And in order to estimate the population variance we can use the sample variance formula:

s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

But we need to calculate the mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_I}{n}

And replacing we got:

\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072

And for the sample variance we have:

s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance \sigma^2

E(s^2) = \sigma^2

3 0
3 years ago
Large Number Calculations
pantera1 [17]

Answer:

19.2 kg

Step-by-step explanation:

Amount of Bananas consumed in US per year = 5.77 million metric tons

Since 1 million = 10^{6} and 1 metric ton = 1000 kg, we can write:

Amount of Bananas consumed in US per year = 5.77 \times 10^{6} metric tons

Amount of Bananas consumed in US per year = 5.77 \times 10^{6} \times 1000 = 5.77 \times 10^{9} kg

Number of people in US = 301 million = 301 \times 10^{6}= 3.01 \times 10^{8}

We have to find how many kilograms of bananas are consumed per person in 1 year in US. For this we have to divide the total amount of bananas eaten in US per year with total number of people in US, which will be:

\frac{5.77 \times 10^{9}}{3.01 \times 10^{8}} \\\\ = 19.2

This means, 19.2 kilograms of bananas are eaten in US per person in a year.

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pav-90 [236]

Answer:

-54 degrees F

Step-by-step explanation:

The coldest temperature ever recorded east of the Mississippi is fifty-four degrees below zero in Danbury, Wisconsin, on January 24, 1922.


The temperature is below zero so it is negative,

-54 degrees F

3 0
3 years ago
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