Simplify: A. 4 B. 11/2 C. 1/4 D. 11/8

According to a survey of adults, 64 percent have money in a bank savings account. If we were to survey 50 randomly selected adul

zheka24 [161]

Answer:

The mean number of adults who would have bank savings accounts is 32.

Step-by-step explanation:

For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

In this problem, we have that:

$p = 0.64$

If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

This is E(X) when $n = 50$ .

So

$E(X) = np = 50*0.64 = 32$

The mean number of adults who would have bank savings accounts is 32.

6 0

4 years ago

Which of the following is the surface area of the right cylinder below?

Norma-Jean [14]

Answer: (A) 352 π units^2
Explanation in image.

Knowledge point:
Geometry - cylinder area formula

6 0

3 years ago

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

5 0

3 years ago

A line passes through the points (-2, 4) and (6, 16). Which of the following would not be

kvasek [131]

(4,12) is the correct answer

3 0

3 years ago

Determine which of the following equations, when graphed, intersect at the point (4, 0). Select all that apply. x - y = 4 -x - y

Elza [17]

Answer: only (x-y=4) and (x+y=4).

Step-by-step explanation:

Notice target point has y=0, so all terms with y are zero. Then all 6 equations reduce to m x = k, for various m and k. So calculate 4×m and compare to k six times.

x - y = 4 4=4 yes

-x - y = 4 -4=4 no

2x - y = 7 8=7 no

x + y = 4 4=4 yes

2x + y = 7 8=7 no

2x + y = -7 8=-7 no.

6 0

3 years ago

Read 2 more answers