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inessss [21]
3 years ago
14

HEY PLS HELP :))))))))))))

Mathematics
1 answer:
nevsk [136]3 years ago
3 0
What do you need help with I got you just send me itnjkjrhhvd
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1. In a karate class there are 22 students. 15 of the students are male. What 6 is the ratio of females to the total number of s
Over [174]

Answer:

15:22

the ratio of men to women in a karate class is 3.3:1. If there are 100 women, how many men are there?Step-by-step explanation:

5 0
3 years ago
If 60% of a number is 18, what is 90%of the number?
TiliK225 [7]
1) B. 16 2) D.14.80 Multiply the percentage by the amount If using a calculator, 90% is 90/100
5 0
4 years ago
The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g and a standard deviation of 0.
babunello [35]

Answer:

a) There is a probability P=0.4992 that a randomly selected candy weights at least 0.8543 g.

b) There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c) The claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Step-by-step explanation:

The average weight of the candies in the package is:

\bar x=\dfrac{\sum x_i}{n}=\dfrac{400.3}{469}=0.8535

For a randomly selected candy (is a sample of size n=1) the standard deviation is the population standard deviation (sigma=0.0511 g).

The probabiltiy that is weights more than 0.8536 can be calculated with the z-score.

z=\dfrac{X-\mu}{\sigma/\sqrt{n}}= \dfrac{0.8536-0.8535}{0.0511/\sqrt{1}}=\dfrac{0.0001}{0.0511}=0.002

P(X>0.8536)=P(z>0.002)=0.4992

There is a probability P=0.4992 that a randomly selected candy weights at least 0.8536 g.

b. If the sample is now of n=441 candies, and we want to know the probability that the mean weight is at least 0.8543 g, the z-score needs to be recalculated:

z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8543-0.8535}{0.0511/\sqrt{441}}=\dfrac{0.0008}{0.0024}=0.3288

P(X_s>0.8543)=P(z>0.3288)=0.37115

There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c. To be more confident about the claim that the mean weight is 0.8556 g, we can calculate the probability that, for a package of 469 candies and using the mean of 0.8535 g that we calculated before, the average weight is at least 0.8556 g.

z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8556-0.8535}{0.0511/\sqrt{469}}=\dfrac{0.0021}{0.0024}=0.89

P(X_s>0.8556)=P(z>0.89)=0.18673

The probability is P=0.187, so the claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

6 0
3 years ago
What is 4 divided by 154
tangare [24]
The answer is 38.5 Your welcome.
6 0
4 years ago
Anyone know the answer thanks a lot
bogdanovich [222]
Y=10 because ~ means that there was a direct dilation. So if you compare 4 & 8, you can tell.
6 0
3 years ago
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