A class' test scores are normally d istributed with a mean of 48 points  and a standard  deviation of 3 points. What percent age of test scores are are between  47 and 52 
points? Round to the nearest hundre dth of a percent.

53.94%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 48 points

σ is the population standard deviation = 3 points

For x = 47 points

z = 47 - 48/3

= -0.33333

Probability value from Z-Table:

P(x = 47) = 0.36944

For x = 52 points

z = 52 - 48/3

= 1.33333

Probability value from Z-Table:

P(x = 52) = 0.90879

Hence, the probability of test scores that between 47 and 52 points is calculated as:

P(x = 52) - P(x = 47)

= 0.90879 - 0.36944

= 0.53935

Therefore, the percentage of test scores that are between 47 and 52

points is given as:

0.53935 × 100

= 53.935%

Approximately to the nearest hundredth of a percent ≈ 53.94%

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950

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step-by-step explanation:

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