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2 years ago

Sam earns $12.50 an hour cleaning houses. If he works from 8:00am to 5:00pm,

Mathematics

2 answers:

lisov135 [29]2 years ago

8 0

Answer:

He would get $112.50.

Step-by-step explanation:

Why?

Because 8 in the morning until 5 in the afternoon is 9 hours. He makes $12.50 an hour, so you multiply 12.50 by 9 and you get the answer...

$112.50

Mumz [18]2 years ago

5 0

8-5 is 9 hours
9hrs x 12.50 = $112.5

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In the DBE 122 class, there are 350 possible points. These points come from 5 homework sets that are worth 10 points each and 3

Sliva [168]

Answer:

i) it is not possible for the student to receive an A in the class

ii) 119 points

iii) 84points

Step-by-step explanation:

Total exam scores = 350points

homework scores of 7, 8, 7, 5, and 8

Let the third exam score =y

Exam scores = 81, 80, x

i) To know if the student would get an A in class, we would find the third exam score

(Scores received by a student)/ (total scores) = least of the grade percentage to get an A

(7 + 8 + 7 +5 + 8 + 81 + 80 + x)/350 = 0.9

(196+x)/350 = 0.9

196+x = 350 × 0.9

196+x = 315

x = 315-196

x = 119

119 > 100

Since the maximum grade for each of the exam score is 100points, it is not possible for the student to receive an A in the class.

ii) Since the least of the grade percentage that would guarantee an A is 0.9, the minimum score on the third exam that will give an A = 119points

iii) (Scores received by a student)/ (total scores) = least of the grade percentage to get a B

(7 + 8 + 7 +5 + 8 + 81 + 80 + x)/350 = 0.8

(196+x)/350 = 0.8

196+x = 350 × 0.8 = 280

x = 280-196

x = 84

The minimum score on the third exam that will give a B = 84points

3 0

3 years ago

Carlos spent 1 1/4 hours doing his math homework. He spent 1/4 of his time practicing his multiplication facts.How many hours di

tamaranim1 [39]

Idk think its 10/4 and simplfy it to 1 and 2/4 1 2/4

3 0

2 years ago

Read 2 more answers

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

2 years ago

According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age grou

Murljashka [212]

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

$p=X/n=35/140=0.25$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239$

We calculate the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198$

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

$P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183$

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

$p=X/n=21/140=0.15$

We can calculate then the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906$

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

$P(X$

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231$

We can calculate the z-scores for p1=0.18 and p2=0.25:

$z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651$

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

$P=P(0.18$

5 0

3 years ago

Solve for the inverse g(x) = 2x^2-6/9

oksian1 [2.3K]

Answer:f(x)=x2−6x−9 f ( x ) = x 2 - 6 x - 9. Replace f(x) f ( x ) with y y . y=x2−6x−9 y = x 2 - 6 x - 9. Interchange the variables. x=y2−6y−9 x = y 2 - 6 y - 9.

Step-by-step explanation:

4 0

2 years ago