How do you solve 6=3+5(y-2)

Answer correctly and fast please.

professor190 [17]

Answer: C

Step-by-step explanation: your fast answer is C.

7 0

3 years ago

Read 2 more answers

The data set shows the numbers of hours students study the week before a test. Find and interpret the mean absolute deviation of

patriot [66]

Step-by-step explanation:

3 + 4 + 5 + 6 + 6 + 7 + 8 + 8 + 9 + 9 / total of numbers or 10 = 65 ÷ 10 = 6.5 This is the mean.

3 - 6.5 = 3.5

4 - 6.5 = 2.5

5 - 6.5 = 1.5

6 - 6.5 = 0.5

7 - 6.5 = 0.5

8 - 0.5 = 7.5

9 - 0.5 = 8.5

MAD = 1.7

4 0

3 years ago

Devaughn is 15 years younger than Sydney. The sum of their ages is 73. What is Sydney’s age?

Olegator [25]

Answer:

Sydney is 44 years old

Step-by-step explanation:

Step 1: Make a system of equations

d = s - 15

d + s = 73

Step 2: Plug in s - 15 for d in the second equation and solve

s - 15 + s = 73

2s - 15 = 73 + 15

2s / 2= 88 / 2

s = 44

Answer: Sydney is 44 years old

6 0

4 years ago

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Al goes 8 mph Ben goes 22 mph in the opposite direction. How many hours has ben traveled when they are 180 miles apart?

NeTakaya

after 6 hours Al and Ben will be 180 miles apart.

(22 * x) + (8 * x) = 180

x = 6 hours

3 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago