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kari74 [83]
4 years ago
13

The sum of 2 numbers is 10 the difference is 6 what are the numbers calculator

Mathematics
1 answer:
IgorLugansk [536]4 years ago
3 0
8 & 2

8+2 = 10

8-2 = 6
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In a randomly selected sample of 1169 men ages 35–44, the mean total cholesterol level was 210 milligrams per deciliter with a s
Aneli [31]

Answer:

The highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10% is 160.59 milligrams per deciliter.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 210, \sigma = 38.6

Find the highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10%.

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

-1.28 = \frac{X - 210}{38.6}

X - 210 = -1.28*38.6

X = 160.59

The highest total cholesterol level a man in this 35–44 age group can have and be in the lowest 10% is 160.59 milligrams per deciliter.

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The prism below is made of cubes which measure 1/3 of a centimeter on one side. What is the volume?
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Answer:

the answer is C.  20/27 cubic cm.

Step-by-step explanation:

V = LWH

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width 2(1/3 cm)

height 2(1/3 cm)

V = (5/3 cm) (2/3 cm) (2/3 cm)

V = 20/27 cubic cm.

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Step-by-step explanation:

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p
Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is  \frac{23}{30} or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

P(A \cup B)=P(A)+P(B)-P(A \cap B)---------------(1)

where

P(A) = Probability of choosing  a student playing basket ball

P(B) =  Probability of choosing  a student playing base ball

P(A \cap B) =  Probability of choosing  a student playing both the games

<u>Finding  P(A)</u>

P(A) = \frac{\text { Number of students playing basket ball }}{\text{Total number of students}}

P(A) = \frac{19}{30}--------------------------(2)

<u>Finding  P(B)</u>

P(B) = \frac{\text { Number of students playing baseball }}{\text{Total number of students}}

P(B) = \frac{12}{30}---------------------------(3)

<u>Finding P(A \cap B)</u>

P(A) = \frac{\text { Number of students playing both games }}{\text{Total number of students}}

P(A) = \frac{8}{30}-----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}

P(A \cup B)= \frac{31}{30} -\frac{8}{30}

P(A \cup B)= \frac{23}{30}

7 0
3 years ago
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