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notsponge [240]

4 years ago

10

Milan spent a total of $70 at the grocery store. Of this amount, he spent $63 on fruit. What percentage of the total did he spen

d on fruit?

2 answers:

Ilia_Sergeevich [38]4 years ago

8 0

63/70 =0.9
0.9 * 100 = 90%

so 90% of the total he spent

bekas [8.4K]4 years ago

7 0

95% Is your answer c:

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Alex73 [517]

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b) 0.023

c) 0.341

d) 0.818

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 515

Standard Deviation, σ = 100

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(score greater than 615)

P(x > 615)

$P( x > 615) = P( z > \displaystyle\frac{615 - 515}{100}) = P(z > 1)$

$= 1 - P(z \leq 1)$

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

$P(x > 615) = 1 - 0.8413 = 0.1587 = 15.87\%$

b) b) P(score greater than 715)

$P(x > 715) = P(z > \displaystyle\frac{715-515}{100}) = P(z > 2)\\\\P( z > 2) = 1 - P(z \leq 2)$

Calculating the value from the standard normal table we have,

$1 - 0.977 = 0.023 = 2.3\%\\P( x > 715) = 2.3\%$

c) P(score between 415 and 515)

$P(415 \leq x \leq 515) = P(\displaystyle\frac{415 - 515}{100} \leq z \leq \displaystyle\frac{515-515}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%$

$P(415 \leq x \leq 515) = 34.1\%$

d) P(score between 315 and 615)

$P(315 \leq x \leq 615) = P(\displaystyle\frac{315 - 515}{100} \leq z \leq \displaystyle\frac{615-515}{100}) = P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.841 - 0.023 = 0.818 = 81.8\%$

$P(315 \leq x \leq 615) = 81.8\%$

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