Cries In didn’t pay attention in class in my lower grades

Mathematics

1 answer:

seraphim [82]3 years ago

4 0

Answer:

a. 32

b. 48

Step-by-step explanation:

Forty percent of the 80 birds were robins.

a. How many of the birds were robins?

b. How many of the birds were not robins?
___________________
Forty percent of the 80 birds that were robins = 40%/100%

To find how many of the birds were robins, use part/whole = part/whole

40%/100% = ?/80

This will tell us how many exact robins were there

To solve, multiply 80 by 40 and then divide all by 100

80 x 40 ÷ 100 = 32

so 32 of the birds were robins

now for b since 32 of the birds were robins out of 80 birds, subtract 32 from 80.

80 - 32 = 48

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Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. 27% of the possible Z values are grea

inessss [21]

Answer:

27% of the possible Z values are greater than 0.613

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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

In this question, we have that:

$\mu = 0, \sigma = 1$

27% of the possible Z values are greater than

The 100 - 27 = 73rd percentile, which is X when Z has a pvalue of 0.73. So X when the z-score is 0.613.

$Z = \frac{X - \mu}{\sigma}$

$0.613 = \frac{X - 0}{1}$

$X = 0.613$

27% of the possible Z values are greater than 0.613

6 0

3 years ago

Find the measure of 0. (To the nearest tenth)<br> A) 28.8<br> B) 33.4<br> C) 61.2<br> D) 63.4

Wewaii [24]

Answer:

Step-by-step explanation:

Hope this helps

4 0

3 years ago

Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought

babymother [125]

Solution:

Number of students in Mr.Skinner's class who brought lunch from home if there are 20 students in the class=12

Fraction of students who brought lunch from home in Mr. Skinner's class= $\frac{12}{20}=\frac{3}{5}$

Number of students in Ms. Cho's class who brought lunch from home if there are 21 students in the class=14

Fraction of students who brought lunch from home in Ms. Cho's class= $\frac{14}{21}=\frac{2}{3}$

As Siloni is using two 15-section spinners to simulate randomly selecting students from each class and predicting whether they brought lunch from home or will buy lunch in the cafeteria.

Number of Congruent sectors in each Spinner=15

So, if we represent students from Mr. Skinner's class who brought lunch from home in Spinner having 15 congruent Sectors = $\frac{9}{15}$