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kenny6666 [7]
3 years ago
8

Cecilia simplified an expression. Her work is shown below.

Mathematics
2 answers:
Alika [10]3 years ago
8 0

Answer:

  • Step 2

Step-by-step explanation:

<u>Given work on simplification</u>

  • (6 and one-half + 2 and three-fourths) minus 1.5 times (4.5 divided by 0.5)
  • (6 1/2 + 2 3/4) - 1.5 (4.5 ÷ 0.5)

<u>Step 1</u> 9 and one-fourth minus 1.5 times (4.5 divided by 0.5)

  • 9 1/4 - 1.5 (4.5 ÷ 0.5)
  • This is ok as first step is solving parenthesis

<u>Step 2</u> 7.75 times (4.5 divided by 0.5)

  • 7.75 (4.5 ÷ 0.5)
  • Incorrect, subtraction should be the last step

Citrus2011 [14]3 years ago
4 0

Answer: second step.

Step-by-step explanation:

you then have to distribute the answer not subtract 9 1/4 from 1.5

you have to do the parenthesis next

4.5/0.5=45/5=9

then

1.5*9= 13.5

then

9.25-13.5 which I think you are capable of solving

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Answer:

a) 37.31 b) 42.70 c) 0.57 d) 0.09

Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., \mu = 35 and \sigma = 6. We know that the probability density function for a normal distribution with a mean of \mu and a standard deviation of \sigma is given by

f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]

in this case we have

f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]

Let X be the random variable that represents a row score, we find the values we are seeking in the following way

a)  we are looking for a number x_{0} such that

P(X\leq x_{0}) = \int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65, this number is x_{0}=37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number  x_{1} such that

P(X\leq x_{1}) = \int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9, this number is x_{1}=42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

P(28\leq X\leq 38)=\int\limits^{38}_{28} {f(x)} \, dx = 0.57

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

P(41\leq X\leq 44)=\int\limits^{44}_{41} {f(x)} \, dx = 0.09

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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3 years ago
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