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Zinaida [17]
2 years ago
7

A mobile phone service provider wants to survey its customers to study privacy concerns and the sharing of their personal inform

ation. They call 5000 randomly selected phone numbers from a database containing the phone number of every customer. If someone selected doesn't answer, they'll attempt calling back up to 2 more times before giving up on reaching that person. They reach 350 customers with this strategy, and 60% of those reached say they are at least "somewhat concerned" about their personal information being shared without their knowledge or consent. Based on this information, and using a 95 percent confidence level, what is the critical value?
Mathematics
1 answer:
Sav [38]2 years ago
8 0

Answer:

The critical value is 1.645.

Step-by-step explanation:

We are given that a mobile phone service provider wants to survey its customers to study privacy concerns and the sharing of their personal information.

They reach 350 customers with this strategy, and 60% of those reached say they are at least "somewhat concerned" about their personal information being shared without their knowledge or consent.

Let p = % of customers who said that they are at least "somewhat concerned" about their personal information being shared without their knowledge or consent.

The test statistics that would be used here is One-sample z test for proportions which means that we will use the z table for finding the critical value.

Also, it is given that a 95 percent confidence level is used; this means that the level of significance = 1 - Confidence level

                                        = 1 - 0.95 = 0.05

Now, in the z table the critical value of z at 0.05 level of significance for one-tailed test is given as 1.645.

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Answer:

3.50x+6.50=y

Step-by-step explanation:

6 0
2 years ago
The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true
Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478

<u>Part B</u>

P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767

<u>Part C</u>

\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971

6 0
2 years ago
Find x- and y-intercepts. Write ordered pairs representing the points where the line crosses the axes. 2x+3y=6
sweet-ann [11.9K]

Answer:

x-intercept: (3, 0)

y-intercept: (0, 2)

Step-by-step explanation:

For a function like:

y = f(x)

The x-intercept is the value of x when y = 0

the y-intercept is the value of y when x = 0

Also remember that an ordered pair is written as (x, y).

In this case we have the equation:

2*x + 3*y = 6

For the x-intercept, we just replace y by zero in the equation, then we get:

2*x + 3*0 = 6

Solving this for x, we get:

2*x = 6

x = 6/2 = 3

Then in this case, the ordered pair for the x-intercept is (3, 0)

For the y-intercept, we just need to replace x by zero in the equation:

2*0 + 3*y = 6

Solving this for y, we get:

3*y = 6

y = 6/3 = 2

Then the ordered pair for the y-intercept is (0, 2)

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2 years ago
Tamara earns 2,050 every month. She spends 65% on the amount that she earns. The rest of the money is equally divided and deposi
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Answer:

Tamara earns $2,050 each month.

She spends 65% of that, so the amount that she spends is:

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Then the amount that she has left is:

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Then she deposits $358.75 per month, so after x months she has:

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x*$358.75 = $2,500

x = $2,500/$358.75 = 6.96

So after 6.96 moths she will have more than $2,500 in one of her accounts, we can round it to:

After 10 months Tamara has deposited more than 2,500 in one of her accounts.

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3 years ago
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