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3 years ago

Noelle has 5/6 of a yard of purple ribbon and 9/10 of a yard of pink ribbon. How much ribbon does she have altogether?

2 answers:

5 0

25/30 + 27/30 = 52/30 = 1 22/30 = 1 11/15 Your answer is A

Hope this helps :)

If you don’t mind, please mark brainliest

Gwar [14]3 years ago

4 0

Answer:

(A) $1\frac{11}{15}$ yards

Step-by-step explanation:

On one hand, Noelle has $\frac{5}{6}$ of a yard of purple ribbon

On the other hand, she has $\frac{9}{10}$ of a yard of purple ribbon.

The length of ribbon she has altogether is determined by the sum of the two fractions.

We add $\frac{5}{6} and \frac{9}{10}$

$\frac{5}{6} + \frac{9}{10}$

Taking the Lowest Common Multiple of 6 and 10 =60

$\frac{5}{6} + \frac{9}{10} =\frac{(5X10)+(9X6)}{60} =\frac{(50)+(54)}{60}=\frac{104}{60}=1\frac{11}{15}$

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Production increased from 8,200 units to 13,530 units; the percent increase was what?

mamaluj [8]

(13530-8200)/8200 times 100

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3 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

$Z = \frac{X - \mu}{s}$

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago

Two consecutive odd integers are such that one-third the smaller is greater than one-seventh of the larger one by 6. Find the nu

vaieri [72.5K]

Answer:

the number is 33 and 35 respectively

Step-by-step explanation:

The computation is shown below:

Let us assume X and X + 2 are the two consecutive odd integers

Now the equation is

(X ÷ 3) = ((X + 2) ÷ 7) + 6

(X ÷ 3) = (X + 2 + 6 × 7) ÷ 7

(X ÷ 3) = (X + 2 + 42) ÷ 7

(X ÷ 3) = (X + 44) ÷ 7

Now do the cross multiplication

7X = 3(X + 44)

7X = 3X + 132

4X = 132

X = 33

And, X + 2 = 33 + 2 = 35

hence, the number is 33 and 35 respectively

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3 years ago

Median of 8,13,17,17,18,22,26,28,33

Andreas93 [3]

Answer:

Median is 18, because 18 is the middle number.

Step-by-step explanation:

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3 years ago

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3 years ago