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3 years ago

15

One of my students got 95 and 98 for the two Principal tests and for the final test. What is the average of her three tests? Wha

t should be the minimal grade on the final test so that the average of the three tests is at least 90?

1 answer:

timofeeve [1]3 years ago

3 0

97% because that is the average of the three test scores.

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The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

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3 years ago

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The world's population has grown at an average rate of 1.9 percent per year since 1945. There were approximately 4 billion peopl

kkurt [141]

Answer: $f(t) = 4000000000( 1 .019)^t$

Step-by-step explanation:

Here, According to the question, The population of the world is 4 billion (approx).

Again According to the question, The world's population has grown at an average rate of 1.9 percent per year since 1945.

Therefore, after 1975 the growth rate will remains same.

Thus, the population of the world after t years,

$f(t) = 4000000000( 1+\frac{1.9}{100} )^t$ (By the formula $A = P(1+\frac{r}{100})^t$ )

⇒ $f(t) = 4000000000( 1.019 )^t$

Where 1.019 is the growth rate and 4 billion is the initial population.

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3 years ago

Please help! Which expression is equal to..?

Alenkasestr [34]

Answer:

A

Step-by-step explanation:

Before we can add the 2 fractions we require them to have a common denominator.

Multiply the numerator/ denominator of the first fraction by x

Multiply the numerator/ denominator of the second fraction by (x + 3)

= $\frac{3x(x)}{x(x+3)}$ + $\frac{(x-2)(x+3)}{x(x+3)}$

Now simplify the numerator, leaving the denominator

= $\frac{3x^2+x^2+x-6}{x(x+3)}$

= $\frac{4x^2+x-6}{x(x+3)}$ → A

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3 years ago

I WILL GOVE YOU BRAINIEST IF YOU CAN SOLVE THIS FIR ME PLEASE!! I need this answered rn!

oee [108]

Answer:

X =2 Y= 10

Step-by-step explanation:

EF ~ BC 8/20= 0.4 AB~DE 15*0.4 =6 6-4 = 2

AC~DF 25*0.4 = 10

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2 years ago