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dedylja [7]
3 years ago
10

In one town, the number of pounds of sugar consumed per person per year has a mean of 8 pounds and a standard deviation 3 of 20

pounds Tyler consumed 13 pounds of sugar last year How many standard deviations from the mean is that? Round your4 answer to two decimal places
A. 250 standard deviations above the mean
B. 2.50 standard deviations below the mean
c. 167 standard deviations above the mean
D. 167 standard deviations below the mean
Mathematics
1 answer:
lesantik [10]3 years ago
3 0

Answer:

c. 1.67 standard deviations above the mean

Step-by-step explanation:

The z-score measures how many standard deviations a score X is from the mean. It is given by the following formula:

Z = \frac{X - \mu}{\sigma}

In which \mu is the mean and \sigma is the standard deviation.

A positive z-score means that X is above the mean, and a negative Z-score means that X i below the mean.

In this problem, we have that:

\mu = 8, \sigma = 3

Tyler consumed 13 pounds of sugar last year How many standard deviations from the mean is that?

This is Z when X = 13. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{13 - 8}{3}

Z = 1.67

So this is 1.67 standard deviations above the mean.

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Answer:

True

Step-by-step explanation:

For example:

HCF of 2 and 3 is 1

5 0
2 years ago
Read 2 more answers
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
Zoe has 4 cucumbers she grew in her garden. She wants to share them equally among 3 of her neighbors.
dedylja [7]

4 divided by 3 = a whole and 3 repeated more than 4 times

(to use the number 4)

:D

8 0
2 years ago
Work out b(4½-3⅔)+1⅔​
Zolol [24]

Answer:

2 \frac{1}{2}

Step-by-step explanation:

Change the mixed numbers to improper fractions

(\frac{9}{2} - \frac{11}{3} ) + \frac{5}{3} ← the LCM of 2 and 3 is 6

= \frac{9(3)}{2(3)} - \frac{11(2)}{3(2)} + \frac{5(2)}{3(2)}

= \frac{27}{6} - \frac{22}{6} + \frac{10}{6}

= \frac{5}{6} + \frac{10}{6}

= \frac{15}{6}

= \frac{5}{2}

= 2 \frac{1}{2}

6 0
2 years ago
In her wallet, Ms. Thompson has one-dollar, five-dollar and ten-dollar bills totaling $171. She has the same number of five-doll
Phantasy [73]
 6 one-dollars = $6 15 five-dollars = $75 9 ten-dollars = $90 Add them all up to get $171, so that is correct. Add the number of one-dollar bills and the number of ten-dollar bills together. 6 + 9 = 15, which is the number of five-dollar bills, so that is correct as well. Add all the numbers of bills together, 6 + 9 + 15 = 15 + 15 = 30.
4 0
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