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3 years ago

Which expression can be used to find the value of expression below -(7/10) + 2/8 + 1/10 - (4/8)

Mathematics

1 answer:

maks197457 [2]3 years ago

4 0

Answer:

(-17)/20

Step-by-step explanation:

Simplify the following:

-7/10 + 2/8 + 1/10 - 4/8

The gcd of 2 and 8 is 2, so 2/8 = (2×1)/(2×4) = 2/2×1/4 = 1/4:

-7/10 + 1/4 + 1/10 - 4/8

The gcd of -4 and 8 is 4, so (-4)/8 = (4 (-1))/(4×2) = 4/4×(-1)/2 = (-1)/2:

-7/10 + 1/4 + 1/10 + -1/2

Put -7/10 + 1/4 + 1/10 - 1/2 over the common denominator 20. -7/10 + 1/4 + 1/10 - 1/2 = (2 (-7))/20 + 5/20 + 2/20 - 10/20:

(2 (-7))/20 + 5/20 + 2/20 - 10/20

2 (-7) = -14:

(-14)/20 + 5/20 + 2/20 - 10/20

-14/20 + 5/20 + 2/20 - 10/20 = (-14 + 5 + 2 - 10)/20:

(-14 + 5 + 2 - 10)/20

-14 + 5 + 2 - 10 = (5 + 2) - (14 + 10):

((5 + 2) - (14 + 10))/20

5 + 2 = 7:

(7 - (14 + 10))/20

| 1 | 4

+ | 1 | 0

| 2 | 4:

(7 - 24)/20

7 - 24 = -(24 - 7):

(-(24 - 7))/20

| 1 | 14

| 2 | 4

- | | 7

| 1 | 7:

Answer: (-17)/20

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A=33

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3 years ago

Mr. O’Brian is paid $7.30 per hour for 40 hours he works in a week. He is paid 1.5 times that rate for each hour after that

posledela

Hey You!

7.30 × 1.5 = 10.95

10.95 × 40 = 438

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3 years ago

WILL GIVE BRAINLIEST 343x^15+125 please show your work

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$343{x}^{15}+125$
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3 years ago

The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1

snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

Answer:

(a). The probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ . (b). The events V and M are not independent.

Step-by-step explanation:

The key factor to solve these questions is to know that:

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

We already know from the question the following probabilities:

$\\ P(V) = 0.73$

$\\ P(M) = 0.18$

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

$\\ P(V \cap M) = 0.03$

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as $\\ P(V \cup M)$ or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

$\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)$

$\\ P(V \cup M) = 0.73 + 0.18 - 0.03$

$\\ P(V \cup M) = 0.88$

Then, the probability that a student has either a Visa card or a MasterCard is $\\ P(V \cup M) = 0.88$ .

Are the events V and M independent?

A way to solve this question is by using the concept of conditional probabilities.

In Probability, two events are independent when we conclude that

$\\ P(A|B) = P(A)$ [1]

The general formula for a conditional probability or the probability that event A given (or assuming) the event B is as follows:

$\\ P(A|B) = \frac{P(A \cap B)}{P(B)}$

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

$\\ P(M|V) = \frac{P(M \cap V)}{P(V)}$

$\\ P(M|V) = \frac{0.03}{0.73}$

$\\ P(M|V) \approx 0.041$

If M were independent from V (according to [1]), we have

$\\ P(M|V) = P(M) = 0.18$

Which is different from we obtained previously;

That is,

$\\ P(M|V) \approx 0.041$

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

$\\ P(V|M)$ , as follows:

$\\ P(V|M) = \frac{P(V \cap M)}{P(M)}$

$\\ P(V|M) = \frac{0.03}{0.18}$

$\\ P(V|M) = 0.1666.....\approx 0.17$

Which is different from

$\\ P(V|M) = P(V) = 0.73$

In the case that both events were independent.

Notice that

$\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)$

$\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03$

$\\ 0.03 = 0.03 = 0.03 = 0.03$

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3 years ago

Jerome is a Pharmacy Technician that filled a prescription that used 1/4 of a bottle of ibuprofen. How many pills did Jerome use

sattari [20]

250

We can use the equation:
f(x)=1/4(x)
f(1,000)=1/4(1,000)
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By using 1/4 of the bottle of ibuprofen, he used 250 pills in each container

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3 years ago