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2 years ago

A bag contains 44 cookies. Kyle eats 8 of the cookies. What fraction of the cookies is left?

Mathematics

2 answers:

NISA [10]2 years ago

5 0

Answer:

9/11

Step-by-step explanation:

36 of the 44 cookies remain. As a fraction, that is 36/44 = 9/11.

Debora [2.8K]2 years ago

4 0

The answer is 36/44 which of we reduce it by dividing both sides by 4 you'll get 9/11
so the answer you'll write down is 9/11

-all you have to do is take the number of cookies in total as the denominator and the numerator is the amount of cookies left when you subtract 44-8=36 And once you get your answer you try to reduce the number to the lowest fraction which in this case is 9/11 :D

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The dot plot below shows how many customers purchased different numbers of shirts at a sale last weekend.

scoray [572]

Answer:

the answer is 1.16 shirts

Step-by-step explanation:

the mean absolute deviation is found by finding the average of the difference between each data point and the mean of the data.

1st...find the mean of the data by adding all the numbers according to the data plotted and dividing the by the numbers listed; which in this case is 10

1 +2+ 2+ 3+ 3+ 3+ 4+ 4+ 5+ 6 = 3.3

mean is 3.3

then find the difference between the mean and each data point

Data Point = 1 2 2 3 3 3 4 4 5 6

Difference from mean = 2.3 1.3 1.3 0.3 0.3 0.3 0.7 0.7 1.7 2.7

Find the average of these differences by adding the (differences from Mean) by 10

2.3 + 1.3 + 1.3 + 0.3 + 0.3 + 0.3 + 0.7 + 0.7 + 1.7 + 2.7

the mean absolute deviations is 1.16 shirts

4 0

2 years ago

Read 2 more answers

The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi

Vesnalui [34]

Answer:

(a) $X\sim N(\mu = 73, \sigma = 16)$

(b) 0.7910

(d) 0.6464

Step-by-step explanation:

Let X = amount of time that people spend at Grover Hot Springs.

The random variable X is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable X is:

$X\sim N(\mu = 73, \sigma = 16)$

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

$P(60$

*Use a z-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0

2 years ago

Cho A =

FromTheMoon [43]

Step-by-step explanation:

uyghouihh9uhu9h9h79h97gouhh,jhkjhouhouhkjhkugiyt75r75r76f8ug87

5 0

1 year ago

Unit 6 homework 2 need help

kodGreya [7K]

∆PQR and ∆SQT are similar, so corresponding sides occur in fixed ratios with one another. This means

SQ / PQ = ST / PR

Solve for x :

8 / ((x + 5) + 8) = (x - 9) / 21

8 / (x + 13) = (x - 9) / 21

168 = (x - 9) (x + 13)

168 = x² + 4x - 117

x² + 4x - 285 = 0

(x + 19) (x - 15) = 0

⇒ x = -19 or x = 15

But x can't be smaller than 9, otherwise ST = x - 9 would be a negative length. So x = 15.

8 0

1 year ago

6 1/5 + (-7 3/5)

GREYUIT [131]

Since you are subtracting a negative you can do it this way. do 7 3/5 - 6 1/5. that leaves you with 1 2/5. so in the end you will have -1 2/5

5 0

1 year ago