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liq [111]
3 years ago
15

Sixth grade 5.17 Solve percent word problems Yus

Mathematics
1 answer:
vovikov84 [41]3 years ago
4 0

big d jit

dic dic dic dic dic dic dic di cidic idc

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Triangle
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Answer:

octagon  enneagon decagon hendeka

Step-by-step explanation:

7 0
3 years ago
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,

a.

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

b.

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

c.

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

d.

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

e.

P(X<5) = P(X≤4)

           = binomial (4, 15, 0.32, 1)

           = 0.4477

⇒P(X<5) = 0.4477

f.

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

3 0
3 years ago
You spin the spinner and flip a coin. Find the probability of the compound event.
Mars2501 [29]
I believe it is 1/5
It would be the answer because getting a 1 out of the 5 possible outcomes would be 1/5
5 0
4 years ago
Read 2 more answers
In a bag with 3 red chips, 2 yellow chips, and 1 green chip, what is the probability of picking 2 yellow chips on the first 2 tr
e-lub [12.9K]

Answer:

D = 2/3

Step-by-step explanation:

take the amount of red chips that we have and put that as the denominator and take the amount of yellow chips that you have and put that as the numerator.

8 0
3 years ago
Estimate the product by rounding 9x54
Blizzard [7]

9*54=486  

10x50=500

So, either 450 or 500.

3 0
3 years ago
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