Can you please help me because I can’t do it right

How much candy at $1.16 a pound should be mixed with candy worth 86 cent a pound in order to obtain a mixture of 60 pounds of ca

vovangra [49]

Let the weights of the two candies be repres. by x and y.

Then x + y = 60, or x = 60 - y

($1.16 / lb) x + ($0.86 / lb) y = ($1.00 / lb) (60 lb) = $60

Then 1.16(60-y) + 0.86y = 60
69.6 - 1.16y + 0.86y = 60 9.6
9.6 = 0.3y Solving for y, y = ------- = 32 lb
0.3

Then x = (60-32) lb = 28 lb

5 0

3 years ago

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Marcus invested $2000 in a bank account earning simple interest. The interest, I, he has after t years at an interest rate r is

Whitepunk [10]

Answer:

4%

Step-by-step explanation:

<h3>Given</h3>

<u>The equation for simple interest:</u>

I = 2000rt

<u>And values</u>

Total = $2160
t = 2
r = ?

<h3>Solution</h3>

<u>Interest amount is</u>

$2160 - $2000 = $160

<u>Solving for annual interest rate</u>

160 = 2000*2r
160 = 4000r
r = 160/4000
r = 0.04
0.04*100% = 4%

Annual interest rate is 4%

6 0

3 years ago

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Situation: to compute the perimeter of a rectangle you find the sum of two times

Alex_Xolod [135]

Answer:

110

Step-by-step explanation:

$30 + 25 \\ = 55 \times 2 \\ = 110$

5 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

PLEASE PLEASE HELP ANYONE I NEED HELP

Maurinko [17]

Answer:

let me make it its equation I have done this but if this will work have a good day

8 0

3 years ago