All categories

4 years ago

Solve the system of equations using the linear combination method. {9x+5y=35 2x+5y=0 Enter your answers in the boxes.

1 answer:

5 0

9x+ 5y=35
2x + 5y=0 |* -1

9x +5y= 35
-2x -5y= 0
-----------------
7x / = 35
x=35:7
x=5

2x+5y=0
2*5+5y=0
10+5y=0
5y=-10
y= -10:5
y=-2

You might be interested in

zane and shelby's grandfather is 5 times as old as the total of their ages. If their grandfather is 70 and Shelby is 6, how many

wel

G=5(z+s)
70=5(z+6)
70=5z+30
40=5z
z=8

Zane is 8

7 0

3 years ago

Jumlah 3 bilangan ganjil berurutan adalah 159, selisih bilangan terbesar dan terkecil adalah

schepotkina [342]

I want to play with you and your family a very happy and prosperous New year to you and your family a very happy and prosperous New year to you and your family a very happy and prosperous and your bucket is filled to get the

6 0

4 years ago

PLwz help thank youuuuuuuuuuuu

MatroZZZ [7]

y = |x|

This is absolute value because it's a v-shape.

4 0

4 years ago

Refrigerators in a random sample of 60 refrigerators, the mean repair cost was $150.00 and the standard deviation was $15.50. Co

worty [1.4K]

Answer:

The 99% confidence interval is given by ($144.677;$155.323)

We are confidenct at 99% that the true mean for the reapir cost of refrigerators is between 144.677 and 155.323.

Step-by-step explanation:

1) Notation and definitions

n=60 represent the sample size

$\bar X= 150$ represent the sample mean

$s=15.5$ represent the sample standard deviation

m represent the margin of error

Confidence =99% or 0.99

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

2) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . The degrees of freedom are given by:

$df=n-1=60-1=59$

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,59)" for $t_{\alpha/2}=-2.66$

"=T.INV(1-0.005,59)" for $t_{1-\alpha/2}=2.66$

The critical value $tc=\pm 2.66$

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=2.66 \frac{15.5}{\sqrt{60}}=5.323$

4) Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$150 - 2.66 \frac{15.5}{\sqrt{60}}=144.677$

$150 + 2.66 \frac{15.5}{\sqrt{60}}=155.323$

The 99% confidence interval is given by ($144.677;$155.323)

We are confidenct at 99% that the true mean for the reapir cost of refrigerators is between 144.677 and 155.323.

7 0

3 years ago

You have 40 different books (20 math books, 15 history books, and 5 geography books). a. You pick one book at random. What is th

svetoff [14.1K]

Answer:

a. $\frac{3}{8}$

b. $\frac{7}{52}$

c. $\frac{95}{156}$

Step-by-step explanation:

Given,

Total books = 40,

Maths books = 20,

History books = 15,

Geography books = 5,

a. If a book is randomly selected,

Then the probability that the book is a history book,

$=\frac{\text{Ways of choosing history book}}{\text{Ways of choosing any one book}}$

$=\frac{^{15}C_1}{^{40}C_1}$

$=\frac{15}{40}$

$=\frac{3}{8}$

b. If two book is randomly selected,

Then the probability that both books are from history,

$=\frac{\text{Ways of choosing two history book}}{\text{Ways of choosing any two book}}$

$=\frac{^{15}C_2}{^{40}C_2}$

$=\frac{\frac{15!}{2!13!}}{\frac{40!}{2!38!}}$

$=\frac{105}{780}$

$=\frac{7}{52}$

c. If two book is randomly selected,

Ways of selecting any two books from different subjects

= maths and history + maths and geography + history and geography,

$=^{20}C_1\times ^{15}C_1+^{20}C_1\times ^{5}C_1+^{15}C_1\times ^{5}C_1$

$=20\times 15 + 20\times 5 + 15\times 5$

$=300 + 100 + 75$

$=475$

Then the probability that both books are from different subjects

$=\frac{\text{Ways of selecting any two books from different subjects}}{\text{Ways of choosing any two book}}$

$=\frac{475}{780}$

$=\frac{95}{156}$

7 0

3 years ago