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3 years ago

112 is 80% of what number?

1 answer:

4 0

112 is 80% of 140.
80% can be written as 0.80 or 0.8.
how i got 140 is by finding the x value of this equation: 0.8x=112
to find the x, i divided 0.8 from both sides,
0.8 divided by 0.8 cancel out.
112 divided by 0.8 is 140.
therefore, 112 is 80% of 140

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4 years ago

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From the figure, we immediately have

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2 years ago

a 9 pound bag of sugar is being split into containers that hold 2/3 of a pound. How many containers of sugar will the 9 pound ba

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4 years ago

Read 2 more answers

What is 0.679 multiplied by 10

motikmotik

Answer:

6.79

Step-by-step explanation:

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3 years ago

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The same survey database cited in exercise 4.3.1 (A-5) shows that 32 percent of U.S. adults indicated that they have been tested

yawa3891 [41]

Answer:

a. P(X = 3) = 0.145736

b. P(5≤X≤9) = 0.5462

c. P(5<X<10) = 0.3332

d. P(X≥6) = 0.3393

e. P(X<5) = 0.4477

f. Mean = 4.8, Variance = 3.264

Step-by-step explanation:

Given - The same survey database cited in exercise 4.3.1 (A-5) shows

that 32 percent of U.S. adults indicated that they have been

tested for HIV at some point in their life. Consider a simple

random sample of 15 adults selected at a time. Let X be the

number of adults who have been tested for HIV in the sample.

To find - For the following, find the numerical answer and describe

the answer in words:

a. Three

b. Between five and nine, inclusive

c. More than five, but less than 10

d. Six or more

e. Less than five

f. Find the mean and the variance of the number of people

tested for HIV in samples of size 15.

Proof -

Given that , n = 15, p = 32% = 0.32

Now,

P(X = 3) = binomial distribution (3, 15, 0.32, 0)

= $\left(\begin{array}{ccc}15\\3\end{array}\right)(0.32)^{3}(1-0.32)^{15 - 3}$

= 455(0.032768)(0.009774)

= 0.145736

⇒P(X = 3) = 0.145736

P(5≤X≤9) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

= $\left(\begin{array}{ccc}15\\5\end{array}\right)(0.32)^{5}(1-0.32)^{15 - 5}$ + $\left(\begin{array}{ccc}15\\6\end{array}\right)(0.32)^{6}(1-0.32)^{15 - 6}$ + $\left(\begin{array}{ccc}15\\7\end{array}\right)(0.32)^{7}(1-0.32)^{15 - 7}$ + $\left(\begin{array}{ccc}15\\8\end{array}\right)(0.32)^{8}(1-0.32)^{15 - 8}$ + $\left(\begin{array}{ccc}15\\9\end{array}\right)(0.32)^{9}(1-0.32)^{15 - 9}$

= 0.213 + 0.1671 + 0.1011 + 0.0476 + 0.0174

= 0.5462

⇒P(5≤X≤9) = 0.5462

P(5<X<10) = + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

= $\left(\begin{array}{ccc}15\\6\end{array}\right)(0.32)^{6}(1-0.32)^{15 - 6}$ + $\left(\begin{array}{ccc}15\\7\end{array}\right)(0.32)^{7}(1-0.32)^{15 - 7}$ + $\left(\begin{array}{ccc}15\\8\end{array}\right)(0.32)^{8}(1-0.32)^{15 - 8}$ + $\left(\begin{array}{ccc}15\\9\end{array}\right)(0.32)^{9}(1-0.32)^{15 - 9}$

= 0.1671 + 0.1011 + 0.0476 + 0.0174

= 0.3332

⇒P(5<X<10) = 0.3332

P(X≥6) = 1 - P(X < 6)

= 1 - P(X ≤ 5)

= 1 - binomial(5, 15, 0.32, 1)

= 1 - 0.6607

= 0.3393

⇒P(X≥6) = 0.3393

P(X<5) = P(X≤4)

= binomial (4, 15, 0.32, 1)

= 0.4477

⇒P(X<5) = 0.4477

Mean = np

= 15(0.32)

= 4.8

⇒Mean = 4.8

Variance = np(1-p)

= 15(0.32)(1 - 0.32)

= 4.8(0.68)

= 3.264

⇒Variance = 3.264

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3 years ago