I NEED HELP WITH (33) B,C,D

Please help me asap!!!!! Worth 20 points

jeka57 [31]

Answer:

Step-by-step explanation

Third, second, first.

The third has 5% chance that it will get the investor its money back, second having a 34% of the investor getting its money back, and the first having a 54% chance. Hope this was quick enough for you!

8 0

1 year ago

Read 2 more answers

What is the solution of the system of equations? y=56x+1 −x+y=2

Rzqust [24]

-x+y=2
y=2+x

2+x=56x+1
x=56x-1
-55x=-1
x=0.181

8 0

3 years ago

You deposit 10,000 into a savings account that compounds quarterly. The account has an interest rate of 3.7%. What would be the

Vadim26 [7]

Answer:

30189.155

Step-by-step explanation:

Recall the compound interest formula :

A = P(1 + r/n)^nt

P = principal = 10000

r = rate = 3.7% = 0.037

n = quarterly = 4

t = 30

A = 10000(1 + 0.037/4)^30*4

A = 10000(1 + 0.00925) ^120

A = 10000(1.00925)^120

A = 10000 * 3.0189155

A = 30189.155

3 0

3 years ago

Write the equation of an ellipse with vertices at (7, 0) and (-7, 0) and co-vertices at (0, 1) and (0, -1).

Irina-Kira [14]

Answer:

$\frac{(x)^{2}}{49}+\frac{(y)^{2}}{1}=1$

Step-by-step explanation:

In this problem we have a horizontal ellipse, because the major axis is the x-axis

The equation of a horizontal ellipse is equal to

$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}} =1$

where

(h,k) is the center of the ellipse

a and b are the respective vertices distances from center

we have

vertices at (7, 0) and (-7, 0)

co-vertices at (0, 1) and (0, -1)

so

The center is the origin (0.0) (The center is the midpoint of the vertices)

a=7

b=1

substitute

$\frac{(x-0)^{2}}{7^{2}}+\frac{(y-0)^{2}}{1^{2}}=1$

$\frac{(x)^{2}}{49}+\frac{(y)^{2}}{1}=1$

8 0

3 years ago

A survey reports that 67% of college students prefer to drink more coffee during the exams week. If we randomly select 80 colleg

Akimi4 [234]

Answer:

The probability that at most 50 say that they drink coffee during exam week is 0.166.

Step-by-step explanation:

The random variable X can be defined as the number of college students who prefer to drink more coffee during the exams week.

The probability of the random variable X is p = 0.67.

A random sample of n = 80 college students are selected.

The response of every students is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 80 and p = 0.67.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np=80\times 0.67=53.6>10\\n(1-p)=80\times (1-0.67)=26.4>10$

Thus, a Normal approximation to binomial can be applied.

$X\sim N(np, np(1-p))$

The mean of the distribution of X is:

$\mu=np=80\times 0.67=53.6$

The standard deviation of the distribution of X is:

$\sigma=\sqrt{np(1-p)}=\sqrt{80\times 0.67\times (1-0.67)}=4.206$

A Normal distribution is a continuous distribution. So, the probability at a point cannot be computed for the Normal distribution. To compute the probability at a point we need to apply the continuity correction.

Compute the probability that at most 50 say that they drink coffee during exam week as follows:

Apply continuity correction:

$P(X\leq 50)=P(X$

$=P(X$

*Use a z-table for the probability.

Thus, the probability that at most 50 say that they drink coffee during exam week is 0.166.

5 0

3 years ago