A student started solving the system of equations by elimination and made a mistake in solving x. What is the error?

Two numbers have the ratio of 4:7 if their sum is 143 what are the numbers?

garri49 [273]

Answer:

52:91

Step-by-step explanation:

To find a ratio value you need to think of it more as two numerators that together equal one whole. 4+7 is 11 so that would be the denominator. If the sum is 143, then we divide by 11 since one number is 4/11s of 143, and the other is 7/11s of 143. 143/11=13

now we multiply both:

4(13)=52

7(13)=91

Check:

91+52=143

52/91 simplifies to 4/7 or 4:7

i hope this helps!

3 0

3 years ago

Michelle and Michael collect baseball cards. Michael has 5 more cards then Michelle. Together they have 27 cards. How many cards

kap26 [50]

Michael is x. Michelle is y.
x+y=27
x-5=y

Substitute y
x+(x-5)= 27
2x-5=27
2x=32
x=16

Plug in x value
(16)+y=27
y=11

Final answer: Michael-16, Michelle-11

5 0

3 years ago

Read 2 more answers

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time coll

algol13

Answer:

(a) The sample size required is 2401.

(b) The sample size required is 2377.

(c) Yes, on increasing the proportion value the sample size decreased.

Step-by-step explanation:

The confidence interval for population proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hatp(1-\hat p)}{n}}$

The margin of error in this interval is:

$MOE=z_{\alpha/2}\sqrt{\frac{\hatp(1-\hat p)}{n}}$

The information provided is:

MOE = 0.02

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

(a)

Assume that the proportion value is 0.50.

Compute the value of n as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\n=\frac{1.96^{2}\times0.50(1-0.50)}{0.02^{2}}\\=2401$

Thus, the sample size required is 2401.

(b)

Given that the proportion value is 0.55.

Compute the value of n as follows:

$MOE=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.55(1-0.55)}{n}}\\n=\frac{1.96^{2}\times0.55(1-0.55)}{0.02^{2}}\\=2376.99\\\approx2377$