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3 years ago

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QUICK HELP!!! Daniel guesses that he scored a 95 on his science test. His actual score is 91. What is the percent error in Danie

ls guess? Round your answer to the nearest tenth of a percent.

2 answers:

bija089 [108]3 years ago

8 0

The answer is %0.8635 hope this helps

deff fn [24]3 years ago

6 0

Answer:

4.4%

Step-by-step explanation:

Because I took the test!

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If f(x) = 7x^2 - 3x +1 and g(x) = 3x - 2, find the following by matching.

Cloud [144]

f(x) = 7x² - 3x + 1

g(x) = 3x - 2

1. g(0) This means that x is 0, so you can plug in 0 for x in the equation:

g(x) = 3x - 2

g(0) = 3(0) - 2

g(0) = -2

2. g(1) x is 1

g(x) = 3x - 2

g(1) = 3(1) - 2

g(1) = 3 - 2

g(1) = 1

3. f(1) x is 1

f(x) = 7x²- 3x + 1

f(1) = 7(1)² - 3(1) + 1

f(1) = 7 - 3 + 1

f(1) = 5

4. f(x) = 7x²- 3x + 1

f(-2) = 7(-2)²- 3(-2) + 1

f(-2) = 7(4) + 6 + 1

f(-2) = 28 + 7

f(-2) = 35

4 0

3 years ago

Answer please I am begging !!!!

Jobisdone [24]

7,134 + d = 10,000

Subtract 7,134 from each side:

d = 10,000 - 7,134

d = 2,866

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4 years ago

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A teacher is making worksheets for the next unit. He can create 3 worksheets in an hour and has 10 worksheets already created fr

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Yuh it’s 37 yuh abcdefcsi

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3 years ago

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Trent had 19 stickers. He bought 29 stickers from a store in the mall and got 18 stickers for his birthday. Then Trent gave 8 of

zlopas [31]

Answer:

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Step-by-step explanation:

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3 years ago

A sample of 16 eggs yields a sample mean weight of 3.00 ounces and a sample standard deviation = 0.360 ounces, find the margin o

shusha [124]

Answer:

$ME =1.753\frac{0.360}{\sqrt{16}}= 0.158$

$3-1.753\frac{0.360}{\sqrt{16}}=2.842$

$3+1.753\frac{0.360}{\sqrt{16}}=3.158$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=3$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.360 represent the sample standard deviation

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,15)".And we see that $t_{\alpha/2}=1.753$

The margin of error is given by:

$ME =1.753\frac{0.360}{\sqrt{16}}= 0.158$

Now we have everything in order to replace into formula (1):

$3-1.753\frac{0.360}{\sqrt{16}}=2.842$

$3+1.753\frac{0.360}{\sqrt{16}}=3.158$

So on this case the 90% confidence interval would be given by (2.842;3.158)

3 0

4 years ago