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Andru [333]
3 years ago
12

Solve the equation -1/3m -7=5

Mathematics
1 answer:
Naddik [55]3 years ago
4 0

-1/3 m - 7 = 5

add 7 to each side

-1/3 m = 12

multiply by -3 to each side

m = -36

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How to solve negative exponents? (Extended,long)
defon
Easily, you write the inverse of it without a minus
so if you were to find x^(-2) it would be (1/x)^2
The rest is all the same
6 0
3 years ago
Jennifer has 25 coins with a total value of $4.25. The coins are quarters and nickels. How many of each does she have?
kirill115 [55]

Answer:

15 quarters, 10 nickels

Step-by-step explanation:

q+n=25

0.25q+0.05n=4.25

multiply the first equation by -0.25 and add the second equation to it

-0.25q-0.25n=-6.25\\0.25q+0.05n=4.25
________________
                -0.2n=-2

n=\frac{-2}{-0.2} =10   has 10 nickels

q=25-n=25-10=15  has 15 quarters

10(.05) +15(0.25) = 0.5 + 3.75 = 4.25

Hope this helps

8 0
1 year ago
Here is a list of numbers 369756707 what is the median?
Korvikt [17]

Answer:

4

Step-by-step explanation:

In statistics and probability theory, the median is the value separating the higher half from the lower half of a data sample, a population or a probability distribution. For a data set, it may be thought of as the "middle" value. And the middle value in this problem is 4.

8 0
3 years ago
Read 2 more answers
A fair coin is tossed three times and the events A, B, and C are defined as follows: A: \{ At least one head is observed \} B: \
Yanka [14]

Answer:

a) P(A)=0.875

b) \text{P(A or B)}=0.875

c) \text{P((not A)  or B  or (not C))}=0.625

Step-by-step explanation:

Given : A fair coin is tossed three times and the events A, B, and C are defined as follows: A: At least one head is observed, B: At least two heads are observed, C: The number of heads observed is odd.

To find : The following probabilities by summing the probabilities of the appropriate sample points ?

Solution :

The sample space is

S={HHH,HHT,HTT,HTH,TTT,TTH,THH,THT}

n(S)=8

A: At least one head is observed

i.e. A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

n(A)=7

B: At least two heads are observed

i.e. B={HHH,HTT,TTH,THT}

n(B)=4

C: The number of heads observed is odd.

i.e. C={HHH,HTT,THT,TTH}

n(c)=4

a) Probability of A, P(A)

P(A)=\frac{n(A)}{n(S)}

P(A)=\frac{7}{8}

P(A)=0.875

b) P(A or B)

Using formula,

\text{P(A or B)}=P(A)+P(B)-\text{P(A and B)}

\text{P(A or B)}=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{\text{n(A and B)}}{n(S)}

\text{P(A or B)}=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}

\text{P(A or B)}=\frac{7}{8}

\text{P(A or B)}=0.875

(c) P((not A)  or B  or (not C))

A={HHH,HHT,HTT,HTH,TTH,TTH,THH,THT}

not A = {TTT} = 1

B={HHH,HTT,TTH,THT}

C={HHH,HTT,THT,TTH}

not C = {HHT,HTH,THH,TTT} = 4

So, not A or B or not C = {HHH,HHT,HTH,THH,TTT}=5

\text{P((not A)  or B  or (not C))}=\frac{5}{8}

\text{P((not A)  or B  or (not C))}=0.625

4 0
3 years ago
Only ARE will answer please what is 4-5=what is the answer
n200080 [17]
Is it -2123 that is my answer
3 0
3 years ago
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