PLEASE HELP QUICK!!! Henry is going to invest in an account paying an interest rate of 4.3% compounded

Helppppppp pls this is due in a few hours

Firdavs [7]

Answer:

Y= -9x+4

Positive x negative y means that it must be negative, then after addition, you get +4.

Step-by-step explanation:

3 0

3 years ago

1 3/10 x 18 please show work and answer

klasskru [66]

Answer:

Step-by-step explanation:

1 3/10 x 18

(1 x 10 +3)/10 = 13/10

13/10 x 18 = (13 x 18) / 10 = 234/10 = (234/2) / (10/2) = 117/5 = 23 2/5 (a fraction is a division)

7 0

3 years ago

Keith is the owner of a cafe. He bought 28.6 pounds of apples for making pies. Apples cost$1.75 per pound. He also bought a can

Blizzard [7]

Answer:

$77.514

Step-by-step explanation:

28.6 x 1.74 + 27.75 = 77.514

8 0

2 years ago

Endpoint: (-9,-1), midpoint: (8,14) given the midpoint and one endpoint of a line segment, find the other endpoint.

jolli1 [7]

Answer:

(25, 29)

Step-by-step explanation:

The midpoint formula can be solved to give an expression for one end point of a line segment in terms of the other end point and the midpoint.

<h3>Formula</h3>

The midpoint (M) of segment AB is found from ...

M = (A +B)/2

Solving for B, we get ...

B = 2M -A

<h3>Application</h3>

For A(-9, -1) and M(8, 14), we find the end point B to be ...

B = 2(8, 14) -(-9, -1)

B = (16 -(-9), 28 -(-1)) = (25, 29)

The other end point is (25, 29.

7 0

2 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$