If there were 20 students in the math 101 class how many of them made an a in the course

Mathematics

1 answer:

ser-zykov [4K]3 years ago

3 0

Answer:

4 students

Step-by-step explanation:

Given that there are 20 student in that Math class and 23% made an A in the course, the number of students that made an A is;

23/100 *20 = 46/10 =4.6

This means exactly 4 students made an A in the class of 20 students.

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In the standard (x,y) coordinate plane, point A has coordinates (−4,0) and point B has coordinates (0,−8). What are the coordina

d1i1m1o1n [39]

$A (-4,0), \ B (0,-8) \\ \\The \ midpoint \ M \ of \ the \ line \ segment \ joining \ the \ points \\ \\ (x _{1}, y _{1}) \ and \ (x _{2}, \ y _{2}) \ is : \\ \\ M=(\frac{x_{1}+x_{2}}{2};\frac{y_{1}+y_{2}}{2})$

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3 years ago

What is the value of f(-4) for the function f(x) = 3/2x-9 ?

oksian1 [2.3K]

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3 years ago

Jamel and his sister received a total of $250 for their birthday. If Jamel received $30 less than his sister, how much did each

alexandr402 [8]

Answer:

Jamel - 110 Sister - 140

Step-by-step explanation:

We are going to work backwards. If Jamel received 30 dollars less than his sister, then the total goes down to $220. Then you divide by 2 to split the money between each equally. Now Jamel has $110 and his sister has $110. But you have to add the original $30 back to his sister's total because she got $30 more. Hence Jamel has $110 and his sister has $140.

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3 years ago

Read 2 more answers

Some sources report that the weights of full-term newborn babies in a certain town have a mean of 8 pounds and a standard devia

Elodia [21]

Answer:

0.682 is the probability that one newborn baby will have a weight within one standard deviation of the mean.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8 pounds

Standard Deviation, σ = 0.6 pounds

We are given that the distribution of weights of full-term newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

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

P(weight between 7.4 and 8.6 pounds)

$P(7.4 \leq x \leq 8.6) = P(\displaystyle\frac{7.4 - 8}{0.6} \leq z \leq \displaystyle\frac{8.6-8}{0.6}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%$