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BaLLatris [955]

2 years ago

5

chris has a bag of sweets. there are more than 20 sweets in the bag he shares his sweets equally between six people chris says,

"the number of sweets in the packet was prime." is he correct? circle: yes or no, explain your answer below.

1 answer:

Likurg_2 [28]2 years ago

3 0

No, this is because a prime number is an uneven number. So it can not be prime.

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While shopping for clothes Tracy spent $38 less than three times what Daniel spent right and solve an equation to find how much

Gnom [1K]

Answer:

y=3x-38

Step-by-step explanation:

We know that Tracy spent $38 in less than 3 times than what Daniel spent. Given this information we can put our eqaution together.

Y=3x-38

Hope this helps! :)

6 0

2 years ago

Suppose babies born in a large hospital have a mean weight of 3316 grams, and a standard deviation of 324 grams. If 83 babies ar

Anuta_ua [19.1K]

Answer: 0.129

Step-by-step explanation:

Let $\overline{X}$ denotes a random variable that represents the mean weight of babies born.

Population mean : $\mu= \text{3316 grams,}$

Standard deviation: $\text{324 grams}$

Sample size = 83

Now, the probability that the mean weight of the sample babies would differ from the population mean by greater than 54 grams will be :

$P(|\mu-\overline{X}|>54)=1-P(\dfrac{-54}{\dfrac{324}{\sqrt{83}}}$

hence, the required probability = 0.129

5 0

2 years ago

4.5x-7=20 i need help whats the answer

Dmitry [639]

Your answer will be 6.

7 0

2 years ago

Read 2 more answers

The national average for mathematics SATs in 2011 was 514 and the standard deviation was approximately 40. a) Within what bounda

blondinia [14]

Answer:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

$\mu = 514 , \sigma =40$

For this case we can use the Chebysev theorem who states that "At least $1 -\frac{1}{k^2}$ of the values lies between $\mu -k\sigma$ and $\mu +k\sigma$ "

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

$0.75 = 1-\frac{1}{k^2}$

If we solve for k we can do this:

$\frac{1}{k^2}= 1-0.75=0.25$

$\frac{1}{0.25}= k^2$

$k^2 =4$

$k =\pm 2$

So then we have at last 75% of the data withitn two deviations from the mean so the limits are:

$Lower = \mu -2\sigma = 514- 2*40=434$

$Upper = \mu +2\sigma = 514 + 2*40=594$

4 0

2 years ago

Eliza likes to make daily events into games of chance. For instance, before she went to buy ice cream at the local ice cream par

sergejj [24]

Answer:

1/6

Step-by-step explanation:

two events need to happen: tutti frutti needs to be shown by first spinner and second spinner needs to show dish

probability of tutti frutti = 1/3

probability of dish = 1/2

probability of both events = 1/3 * 1 /2 = 1/6

6 0

2 years ago