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3 years ago

What is 1 1/4 + (3 2/3 +5 3/4)?

Mathematics

1 answer:

n200080 [17]3 years ago

6 0

Answer:

Step-by-step explanation:

Because this whole problem focuses on the addition of mixed numbers, we can drop the parentheses right now without affecting the problem solution.

The two denominators shown are 4 and 3; the LCD is therefore 12.

Rewriting the problem, we get:

1 3/12 + 3 8/12 + 5 9/12, or

3 + 8 + 9

1 + 3 + 5 + ----------------

This simplifies to 9 20/12, or 10 8/12, or 10 2/3 (answer)

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For women aged 18-24, systolic blood pressures are normally distributed with a mean of 114.8 mm Hg and a standard deviation of

NNADVOKAT [17]

Answer:

0.0579 is the probability that mean systolic blood pressure is between 119 and 122 mm Hg for the sample.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 114.8 mm Hg

Standard Deviation, σ = 13.1 mm Hg

Sample size = 23

We are given that the distribution of systolic blood pressures is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{13.1}{\sqrt{23}} = 2.731$

P(blood pressure is between 119 and 122 mm Hg)

$P(119 \leq x \leq 122) = P(\displaystyle\frac{119 - 114.8}{2.731} \leq z \leq \displaystyle\frac{122-114.8}{2.731}) = P(1.537 \leq z \leq 2.636)\\\\= P(z \leq 2.636) - P(z < 1.537)\\= 0.9958 - 0.9379 = 0.0579= 5.79\%$

$P(119 \leq x \leq 122) = 5.79\%$

0.0579 is the probability that mean systolic blood pressure is between 119 and 122 mm Hg for the sample.

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2 years ago

What is the answer to this slope A (1,3),B (4,7)

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3 years ago

Mrs. Daly's reading classes were surveyed about their reading preferences. The data is summarized in the table below.

andriy [413]

Answer:

Required probability is 0.19

Step-by-step explanation:

The two-way table for the given data is,

Fiction Non-Fiction Total

9th grade 35 8 43

10th grade 52 12 64

Total 87 20 107

It is required to find the probability of the students who prefer to read non-fiction novels, given that they are in 9th grade.

That is, to find the conditional probability $P(Non-fiction| 9th\ grade)$ .

Now, $P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}$

So, we get,

$P(Non-fiction| 9th\ grade)=\dfrac{(Non-fiction\bigcap 9th\ grade)}{P(9th\ grade)}\\\\P(Non-fiction| 9th\ grade)=\dfrac{8}{43}\\\\P(Non-fiction| 9th\ grade)=0.19$

<h3>Thus, the required probability is 0.19.</h3>

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3 years ago