Ask question

Ask question

All categories

3 years ago

5

Mr. Miller purchased two coats that cost the same amount from an online store. The total amount Mr. Miller paid was$118. What wa

s the price of one coat?

1 answer:

earnstyle [38]3 years ago

8 0

Answer:

The price of 1 coat is $59.

Step-by-step explanation:

Given that 2 coats have the same price so in order to find the price of 1 coat, you have to divide by 2 :

2 coats = $118

1 coat = $118 ÷ 2

1 coat = $59

You might be interested in

15 points<br>Steps needed

morpeh [17]

It is not a negative so it is 10 because the number can't be big

3 0

3 years ago

Directions: Graph each function and give its key characteristics. Use a graphing calculator for the turning points and round to

skad [1K]

Answer:

for example, if f(x)= x^2 + 3

and they gave write f(5)= 5^2 +3

=25+3=28

3 0

2 years ago

spin [16.1K]

<span>Area of circular metal=πr^2 =3.14 x 3.5 x 3.5 m^2 now, total cost=area x rate =3.14 x 3.5 x 3.5 x 3.25 $ =125.01 $</span>

4 0

3 years ago

Read 2 more answers

) One year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million. Suppose a samp

Nutka1998 [239]

Answer:

Probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

Step-by-step explanation:

We are given that one year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million.

Suppose a sample of 400 major league players was taken.

<em>Let </em> $\bar X$ <em> = sample average salary</em>

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = mean salary = $1.5 million

$\sigma$ = standard deviation = $0.9 million

n = sample of players = 400

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the average salary of the 400 players exceeded $1.1 million is given by = P( $\bar X$ > $1.1 million)

P( $\bar X$ > $1.1 million) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ > $\frac{ 1.1-1.5}{{\frac{0.9}{\sqrt{400} } }} }$ ) = P(Z > -8.89) = P(Z < 8.89)

<em>Now, in the z table the maximum value of which probability area is given is for critical value of x = 4.40 as 0.99999. So, we can assume that the above probability also has an area of 0.99999 or nearly close to 1.</em>

Therefore, probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.

4 0

3 years ago

The director of a state agency believes that the average starting salary for clerical employees in the state is less than $35 c

MrRissso [65]

Answer:

There is not enough evidence to conclude that average starting salary for clerical employees in the state is less than $35,000 per year.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 35,000 dollars per year

Sample mean, $\bar{x}$ = 34,700 dollars per year

Sample size, n = 100

Alpha, α = 0.10

Population standard deviation, σ = 3,500 dollars per year

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 35,000\text{ dollars per year}\\H_A: \mu < 35,000\text{ dollars per year}$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{34700 - 35000}{\frac{3500}{\sqrt{100}} } = -0.8571$

Now, $z_{critical} \text{ at 0.10 level of significance } = -1.28$

Since,

$z_{stat} > z_{critical}$

We fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is not enough evidence to conclude that average starting salary for clerical employees in the state is less than $35,000 per year.

3 0

3 years ago