All categories

3 years ago

10 times as many as _ hundreds is 60 hundreds or _ thousands

Mathematics

1 answer:

Lady_Fox [76]3 years ago

6 0

60 hundreds means 6000 = 6 thousands;

So, 10 · x = 6000;

x = 600;

600 means 6 hundreds;

Finally, 10 times as many as 6 hundreds is 60 hundreds or 6 thousands.

You might be interested in

Work the following pricing problems for services rendered. (For all calculations use hundredths.)

Lesechka [4]

Answer:

Hourly Rate = $12.22 per hour

Step-by-step explanation:

$\text{Labor Time = }4\tfrac{1}{2}=4.5\thinspace{ hours}$

Overhead Rate = 95%

Retail price of parts = $134.75

Total cost of job = $241.98

Cost of labor = Total cost - Retail price

= 241.98 - 134.75

= $107.23

Now, labor works for 4.5 hours and ears a total amount of $107.23

$\implies\text{Amount earned in one hour or hourly rate = }\frac{107.23}{4.5}\\\\\implies \text{Hourly rate = }\$23.83\text{ per hour}$

Now, Labor will get 95% overhead rate

⇒ The labor is getting 100% already but an overhead of 95% is also given

$\text{So, Hourly Rate = }\frac{23.83}{1+0.95}\\\\\bf\implies\textbf{Hourly Rate = }\$12.22\textbf{ per hour}$

3 0

3 years ago

A catering company loses $110 as a result of a delay.

tatyana61 [14]

Answer:What you have to do first is you have 5 owners correct? Well you take anything they're sharing in this case it's delay/debt, so you divide $110 by 5 and you should get $22. This means that each owner would have to pay $22 for the delay.

ANS -22

Brainly please

4 0

3 years ago

write the equation of the line using the given information. type your awnser in slope intercept form,y=mx+b

faltersainse [42]

Answer:

y=3/-1x+4

Step-by-step explanation:

Okay so first thing you need to know is that if there is an ordered pair (x, y) where x is 0, your y-intercept is your y. For example, your problem has (0, 4) your x is 0 and your y is 4. Therefore your y-intercept is 4 which is the b. To find your mx, or slope, you need to do (y2-y1)/(x2-x1). Your y2 will be your y in your second ordered pair and your y1 will be in your y in the first ordered pair. Same for your x. So, (4-1)/(0-1) which equals 3/-1. 3/-1 is your slope. So, your answer in slop intercept form is: y= 3/-1x+4. You could also try y= -3x+4 if that makes you more comfortable.

I know this is long this is my first time doing this lol.

5 0

3 years ago

3. Continue the pattern with 4 fifths.

Elenna [48]

Answer:

I don’t have enough context clues to answer your question I’m afraid

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

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

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

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

2 years ago