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2 years ago

Can someone answer this please? a 22.5mg b 33.75mg c 45mg d 11.25mg

Mathematics

2 answers:

atroni [7]2 years ago

3 0

I think the right answer is b

tester [92]2 years ago

3 0

Answer:

Step-by-step explanation:

to solve, we need to first figure out how much is %75 of 45, an easy way to figure this out is to find the common denominator in both sides, since we know that %75 of something is also 3/4, we can divide by the denominator

45 / 4 = 11.25

now thats 1/4, but thats only %25, so all we need to do now is multiply by how many pieces we have, in this case 3 from 3/4

11.25 * 3 = 33.75

hope this helps you out with your homework :D

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PLS HELP What is the median of this data set?

balu736 [363]

Answer:

$3.50

Step-by-step explanation:

add up all the x's and distribute them starting from the outside until you get to the middle. if there is an odd number of x's (which there isn't), the last x to be placed will be on the spot that is the median. If there is an even number of x's, and they both land on the same spot, that spot will be the median, and if they are next to each other, the median will be between them.

7 0

2 years ago

What is the y intercept, WILL GIVE 15 POINTS WHOEVER GIVES CORRECT ANSWER

DENIUS [597]

Answer:

The relationship between x and y is 6.

Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers

Bob and mark talk about their families. Bob says he has 3 kids, the product of their age is 72. He gives another clue: the sum o

Bingel [31]

Possible Ages Sum of The Ages 1 2 36 391 4 18 231 8 9 181 3 24 282 3 12 172 4 9 152 6 6 142 2 18 223 3 8 143 4 6 13

The sum has to be 14. So it can be 2, 6, and 6 or 3, 3, and 8.
But it cannot be 3, 3, and 8 because the youngest cannot be a twin.

So it is 2, 6, and 6.

Hope this helps :)

3 0

3 years ago

A new sample of 225 employed adults is chosen. Find the probability that less than 7.1% of the individuals in this sample hold m

arsen [322]

Answer:

The probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

Step-by-step explanation:

Let X = number of individuals in the United States who held multiple jobs.

The probability that an individual holds multiple jobs is, p = 0.13.

The sample of employed individuals selected is of size, n = 225.

An individual holding multiple jobs is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 225 and p = 0.13.

But since the sample size is too large Normal approximation to Binomial can be used to define the distribution of proportion p.

Conditions of Normal approximation to Binomial are:

np ≥ 10
n (1 - p) ≥ 10

Check the conditions as follows:

$np=225\times 0.13=29.25>10\\n(1-p)=225\times (1-0.13)=195.75>10$

The distribution of the proportion of individuals who hold multiple jobs is,

$p\sim N(p, \frac{p(1-p)}{n})$

Compute the probability that less than 7.1% of the individuals in this sample hold multiple jobs as follows:

$P(p$

*Use a z-table.

Thus, the probability that less than 7.1% of the individuals in this sample hold multiple jobs is 0.0043.

7 0

3 years ago

PLEASE help me with this.

alexdok [17]

Answer:

I believe C is the line of best fit because everything is spread out more evenly instead of some being spread out farther than others.

3 0

2 years ago