78.84 is what percent of 292? Pleasee, №17887766, 02.01.2020 17:56

Mathematics

78.84 is what percent of 292? Pleasee

Step-by-step answer

P Answered by PhD

1

27%

Step-by-step explanation:

Mathematics

Write the following percents as both fractions and decimals. fraction decimal 1. 25% ? ? 2. 32.5% ? ? 3. 4% ? ? 4. 75% ? ? 5. 6.5% ? ? 6. 125% ? ? 7. 125.5 ? ? 8. 0.2% ? ? 9. 0.75% ? ? 10. 107% ? ? 11. 210% ? ? 12. 22.5% ? ? find the rate, or percent, for each of the following items. round your answers to the nearest 0.1%. base 72; percentage 12 percentage 28; base 224 base 20; percentage 40 base 44; percentage 99 percentage 126; base 8400 find the base for the following items. 50 is 20% of what number? 50 is 20% of what number? 18 is 331⁄3% of

Step-by-step answer

P Answered by Specialist

10

fractions and decimals

1. 25% = 25/100 and 0.25

2. 32.5% = 325/1000 and 0.325

3. 4% = 4/100 and 0.04

4. 75%= 75/100 and 0.75

5. .5% = 65/1000 and 0.065

6. 125% = 125/100 and 1.25

7. 125.5% = 1255/1000 and 1.255

8. 0.2% = 2/1000 and 0.002

9. 0.75%= 75/10000 and 0.0075

10. 107% = 107/100 and 1.07

11. 210%= 210/100 and 2.1

12. 22.5%= 225/1000 and 0.225

Step-by-step explanation:

Mathematics

The number 84 is decreased to 78. What is the percentage by which the number was decreased, to the nearest tenth of a percent?

Step-by-step answer

P Answered by PhD

4

Step-by-step explanation:

it went down by 7.14%

pls mark me brainliest

Mathematics

Camillo estimates the weight of his dog. The veterinarian says that the dog weighs 84 pounds. The percent error in Camillo s estimates is less than 10%. Decide whether each of the following weight estimates could have been Camillo s estimate. a. 80 pounds yes or no b. 78 pounds yes or no c. 94 pounds yes or no d. 77 pounds yes or no e. 92 pounds yes or no f. 75 pounds yes or no

Step-by-step answer

P Answered by Master

12

A.Yes B.Yes C.No D.Yes E.No F.No

Step-by-step explanation:

10% of 84 is 8.4 and 84-8.4=76.6 and anything between 76.6-84 could have been the estimate.

Mathematics

Camillo estimates the weight of his dog. The veterinarian says that the dog weighs 84 pounds. The percent error in Camillo s estimates is less than 10%. Decide whether each of the following weight estimates could have been Camillo s estimate. a. 80 pounds yes or no b. 78 pounds yes or no c. 94 pounds yes or no d. 77 pounds yes or no e. 92 pounds yes or no f. 75 pounds yes or no

Step-by-step answer

P Answered by Master

12

A.Yes B.Yes C.No D.Yes E.No F.No

Step-by-step explanation:

10% of 84 is 8.4 and 84-8.4=76.6 and anything between 76.6-84 could have been the estimate.

Mathematics

Jim tree sells trees. the mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. jim wants to know what percent of his sales were more than 84 inches tall. he can use the standard normal distribution to him. 1. he asks, what is the mean? mean = inches. he thinks, what is the expected distribution about the mean? ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches. ) 2. he thinks, % of the sales will be below 68 inches. but, he needs to know the percentage between 68 and 84 to add to that.so,

Step-by-step answer

P Answered by PhD

13

1. The mean is given in the problem statement as 68 inches.

2. 50% of the normal distribution is below the mean. This question answers itself regarding the z-score. It is 1.6. (Jim computed it as z = (x-μ)/σ = (84-68)/10 = 1.6.)

3. We don't have your table, so we don't know what it tells you. We expect it will tell you either 0.05480 or 0.94520. Assuming the latter, the percentage associated with 1.6 in the table is 94.5%. Now Jim knows 94.5% of his sales were 84 inches or less.

4. Therefore, the remaining 5.5% were more than 84 inches.

The numbers more than and less than a given value must total 100%.

Mathematics

Jim tree sells trees. the mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. jim wants to know what percent of his sales were more than 84 inches tall. he can use the standard normal distribution to him. he asks, what is the mean? mean = inches. he thinks, what is the expected distribution about the mean? ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches. ) he thinks, % of the sales will be below 68 inches. but, he needs to know the percentage between 68 and 84 to add to that.so, he calculates

Step-by-step answer

P Answered by Master

6

Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to know what percent of his sales were more than 84 inches tall. He can use the standard normal distribution to help him.

He asks, "what is the mean?" Mean = 68 inches.

He thinks, "What is the expected distribution about the mean?" ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches.)

He thinks, " 50% of the sales will be below 68 inches."

But, he needs to know the percentage between 68 and 84 to add to that. So, he calculates a z-score
z-score=(x-mean)/(standard deviation)=(84-68)/10=16/10
z-score=1.6

and finds the percentage associated with 1.6 in the table.
0.35543=0.35543*100%=35.543%=35.5%

This is (to the nearest tenth) 35.5%.

Now Jim knows that 50%+35.5%=85.5% of his sales were 84 inches or less.

Therefore, the remaining 100%-85.5%=14.5% were more than 84 inches.

Mathematics

Jim tree sells trees. the mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. jim wants to know what percent of his sales were more than 84 inches tall. he can use the standard normal distribution to him. 1. he asks, what is the mean? mean = inches. he thinks, what is the expected distribution about the mean? ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches. ) 2. he thinks, % of the sales will be below 68 inches. but, he needs to know the percentage between 68 and 84 to add to that.so,

Step-by-step answer

P Answered by Specialist

7

Step-by-step explanation:

We know that,

$Z=\dfrac{X-\mu}{\sigma}$

Where,

X = raw score = 84

μ = mean = 68

σ = standard deviation = 10

Putting the values,

$Z=\dfrac{84-68}{10}=1.6$

The percent of his sales that were less than or equal to 84 inches, is

$P(Z\leq 1.6)=0.9452=94.52\%\approx 94.5\%$

Therefore, the remaining 100-94.5=5.5% were more than 84 inches.

Mathematics

Jim tree sells trees. the mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. jim wants to know what percent of his sales were more than 84 inches tall. he can use the standard normal distribution to him. he asks, what is the mean? mean = inches. he thinks, what is the expected distribution about the mean? ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches. ) he thinks, % of the sales will be below 68 inches. but, he needs to know the percentage between 68 and 84 to add to that.so, he calculates

Step-by-step answer

P Answered by Specialist

6

Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to know what percent of his sales were more than 84 inches tall. He can use the standard normal distribution to help him.

He asks, "what is the mean?" Mean = 68 inches.

He thinks, "What is the expected distribution about the mean?" ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches.)

He thinks, " 50% of the sales will be below 68 inches."

But, he needs to know the percentage between 68 and 84 to add to that. So, he calculates a z-score
z-score=(x-mean)/(standard deviation)=(84-68)/10=16/10
z-score=1.6

and finds the percentage associated with 1.6 in the table.
0.35543=0.35543*100%=35.543%=35.5%

This is (to the nearest tenth) 35.5%.

Now Jim knows that 50%+35.5%=85.5% of his sales were 84 inches or less.

Therefore, the remaining 100%-85.5%=14.5% were more than 84 inches.

History

Analyze the chart below and answer the question that follows. A 5-column table with 4 rows. Column 1 has entries Libya, Tunisia, Egypt, Morocco. Columns 2 and 3 are labeled Literacy Rate (all adults). Column 2 has entries Men, 92.4 percent, 83.4 percent, 71.4 percent, 65.7 percent. Column 3 has entries Women, 72 percent, 65.3 percent, 59.4 percent, 39.6 percent. Columns 4 and 5 are labeled Literacy Rate (15 to 24 years old). Column 4 has entries Men, 99.9 percent, 98.1 percent, 87.9 percent, 84.8 percent. Column 5 has entries 99.7 percent, 95.7

Step-by-step answer

P Answered by PhD

The answer is B.

Explanation:

78.84 is what percent of 292?
Pleasee

Step-by-step answer

Faq

Try asking the Studen AI a question.