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vodka [1.7K]
2 years ago
7

Express 0.92 as a percent​

Mathematics
2 answers:
Andru [333]2 years ago
8 0

Answer:

0.92 as a percent is 92%

Step-by-step explanation:

alekssr [168]2 years ago
7 0
92% in school I learned that in a decimal the zero means nothing so when you take the zero off your left with 92 and you convert it into a percentage therefore your answer is 92%
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Can someone help me thankyou <3
grigory [225]

Answer:

P(G)= 7/10

B, 1, P(not B)

1- 8/10, p(Y)= 2/10

Step-by-step explanation:

hope this helps

correct me if this is wrong

6 0
3 years ago
What is the slope of the line that passes through (-3,-7) and (1,9)
Rus_ich [418]

To find slope, use the formula:

(y2-y1)/x2-x1)

So here is what it will look like:

(9--7)/(1--3)

Since you are subtracting a negative, then it turns unto a positive so you are adding the numbers together.  

(9+7)/1+3)

16/4

4

The slope of the line is 4/1 or 4.  Which means that you will go up or "rise" 4 and go to the left or "run" 1.


Hope this helps!

4 0
3 years ago
Need help on this math problem of parallelograms.
tamaranim1 [39]

Answer:

z=124 because it opposite

Step-by-step explanation:

6 0
3 years ago
How did the temperature change if: at first it decreased by 60% and then increased by 80%. PLEASE NEED HELP!
Sergio039 [100]

The temperature decreased by 28%

Step-by-step explanation:

How did the temperature change if :

  • At first it decreased by 60%
  • Then increased by 80%

Assume that the temperature before decreasing is 100%

∵ The temperature before decreasing = 100%

∵ It decreased by 60%

→  Before decreasing  :  decreasing  :    after decreasing

→  100%                         :  60%              :    100% - 60% = 40%

∴ The temperature after decreasing = 40%

Assume that the temperature after decreasing is 100%

∵ The temperature after decreasing = 100%

∵ It increased by 80%

By using the ratio method

→  After decreasing  :  increasing  :    after increasing

→  100%                      :  80%             :   100% + 80% = 180%

→  40%                        :                       :   x%

By using cross multiplication

∴ 100% × x% = 40% × 180%

- Divide both sides by 100%

∴ x% = \frac{(40)(180)}{100}

∴ x% = 72%

∵ At first the temperature = 100%

∵ At last the temperature = 72%

∵ The first the temperature > the last the temperature

∴ The decreasing in temperature = 100% - 72% = 28%

The temperature decreased by 28%

Learn more:

You can learn more about ratio in brainly.com/question/2707032

#LearnwithBrainly

5 0
3 years ago
Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r
ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

                                        = [ 21.88-1.796 \times {\frac{0.201}{\sqrt{12} } } , 21.88+1.796 \times {\frac{0.201}{\sqrt{12} } } ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0
3 years ago
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