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Stells [14]
3 years ago
5

Simplify the - 501/2+12.3

Mathematics
1 answer:
SOVA2 [1]3 years ago
4 0

Answer:

- 50 \frac{1}{2}  + 12.3 \\  \\  =  \frac{ - 101}{2}  + 12.3 \\  \\  =  - 38.2 \\  \\  =  - 38 \frac{1}{5}

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Fifty-four is ___% of 60.<br><br> 20 points offered
prisoha [69]

Answer:

54÷60×100= 90%

Step-by-step explanation:

90%

cmon man

4 0
3 years ago
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A box contains 512 grams of cereal. one serving is 56 grams. how many servings of cereal dose the box contain?​ GETS BRAINLIEST
kenny6666 [7]

Answer:

the answer is 9.14 so that means about nine servings of cereal

Step-by-step explanation:

you you divide 512 and 56 and you get a total of 9.14

8 0
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Which statement is true of z11.7?
Alexxx [7]

Answer

z11.7 is between 1 and 2 standard deviations of the mean.

Step-by-step explanation:

7 0
3 years ago
Assume the readings on thermometers are normally distributed with a mean of 0degreesc and a standard deviation of 1.00degreesc.
Tanzania [10]

Let Z be the reading on thermometer. Z follows Standard Normal distribution with mean μ =0 and standard deviation σ=1

The probability that randomly selected thermometer reads greater than 2.07 is

P(z > 2.07) = 1 -P(z < 2.07)

Using z score table to find probability below z=2.07

P(Z < 2.07) = 0.9808

P(z > 2.07) = 1- 0.9808

P(z > 2.07) = 0.0192

The probability that a randomly selected thermometer reads greater than 2.07 is 0.0192

3 0
3 years ago
In a recent study on world​ happiness, participants were asked to evaluate their current lives on a scale from 0 to​ 10, where 0
uysha [10]

Answer:

a) A response of 8.9 represents the 92nd ​percentile.

b) A response of 6.6 represents the 62nd ​percentile.

c) A response of 4.4 represents the first ​quartile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5.9

Standard Deviation, σ = 2.2

We assume that the distribution of response is a bell shaped distribution that is a normal distribution.

Formula:

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

a) We have to find the value of x such that the probability is 0.92

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.92

Calculation the value from standard normal z table, we have,  

P(z

\displaystyle\frac{x - 5.9}{2.2} = 1.405\\x = 8.991 \approx 8.9

A response of 8.9 represents the 92nd ​percentile.

b) We have to find the value of x such that the probability is 0.62

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.62

Calculation the value from standard normal z table, we have,  

P(z

\displaystyle\frac{x - 5.9}{2.2} = 0.305\\x = 6.571 \approx 6.6

A response of 6.6 represents the 62nd ​percentile.

c) We have to find the value of x such that the probability is 0.25

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.25

Calculation the value from standard normal z table, we have,  

P(z

\displaystyle\frac{x - 5.9}{2.2} = -0.674\\x = 4.4172 \approx 4.4

A response of 4.4 represents the first ​quartile.

4 0
3 years ago
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