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Karo-lina-s [1.5K]
3 years ago
5

0.667 in simple fraction​

Mathematics
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

667/1000

Step-by-step explanation:

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(PLEASE HELP!!) Use the following diagram to answer the questions.
monitta

Answer:

x:

5x - 29 = 3x+ 19

2x = 48

x = 24°

Angle 1:

180 - (3x+7) = 180 - 79 = 101°

Angle 2:

3x + 7 = 79°

Angle 3:

Same as Angle 1 = 101°

Angle 4:

Same as Angle 1 = 101°

Angle 5:

Same as Angle 2 = 79°

Angle 6:

Same as Angle 5 = 79°

Angle 7:

180 - (5x - 29) = 180 - 91 = 89°

Angle 8:

Same as Angle 7 = 89°

Angles 2 and 3 are Supplementary angles

5 0
3 years ago
Will mark brainliest
vovikov84 [41]
(0,0)-->(0,0)
(3,-1)-->(9,-3)
(3,3)-->(9,9)


so what you will plug after the dilation will be the points on the right/
hope i help
 
3 0
3 years ago
Read 2 more answers
A shelter has enough food to feed an average of 80 hungry people a day for 14 days. If an average of 32 people are at the shelte
egoroff_w [7]

Multiply 80 by 14 to get how many total people they can feed.

80 times 14 is 1120.

Now, divide 1120 by 32 to get how many days the food will last if 32 people come every day.

1120/32 = 35.

The answer is B, 35 days

4 0
4 years ago
Read 2 more answers
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

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)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

6 0
3 years ago
Jessie paid $43.75 for 5 movie tickets. Each ticket cost the same amount. What was the cost of each movie ticket in dollars and
N76 [4]
If I’m not wrong each ticket was 8.75
Because 43.75 divided by 5= 8.75:)
5 0
3 years ago
Read 2 more answers
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