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Paha777 [63]
3 years ago
5

Will give Brainliest!

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
7 0
The answer to this is c. y=2x-6
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Knowing that QPT = ARZ, a congruent side pair is?
vivado [14]

Answer:

\large\boxed{\overline{QT}\cong\overline{AZ}}

Step-by-step explanation:

\triangle QPT\cong\triangle ARZ\\\\\begin{array}{c|c}Q&A\\P&R\\T&Z\end{array}\Rightarrow\begin{array}{ccc}\overline{QP}\cong\overline{AR}\\\overline{PT}\cong\overline{RZ}\\\overline{QT}\cong\overline{AZ}\end{array}

7 0
3 years ago
Read 2 more answers
A coin has a diameter of about 30 millimeters. What is the area of the coin
SOVA2 [1]
Hi the answer is706.5mm
7 0
3 years ago
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A game has three possible outcomes: win, lose, or tie. The probability of a win is 0.4, the probability of a loss is 0.1, and th
kompoz [17]
The probability of not winning will probably be  0.1




THE ANSWER IS AFDHYH7UYHJ\J






 
5 0
2 years ago
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A taxi company charges $4.00 for the first mile (or part of a mile) and 80 cents for each succeeding tenth of a mile (or part).
svet-max [94.6K]

Answer:

The piecewise function is:

C(x) = C(x) = \left \{ {{4, 0< x \leq 1} \atop {4 + 8x, 1 < x \leq 2}}\right.

Step-by-step explanation:

A piecewise function is a function that is defined in multiple intervals.

In the first interval:

0 < x \leq 1

The problem states that a taxi company charges $4.00 for the first mile (or part of a mile).

x is the number of miles. So

If x \leq 1, C(x) = $4.00.

Second interval:

1 < x \ leq 2

Here, the cost is defined by a linear function in the following format:

C(x) = C_{0} + rx

In which C_{0} is the initial price and r is the price paid per mile.

The problem states that each succeeding tenth of a mile costs 80 cents. So

we have the following rule of three.

1 mile - r dollars

0.1miles - 0.8 dollars

0.1r = 0.8

r = \frac{0.8}{0.1}

r = 8

So, we have

C(x) = 4 + 8x, 1 < x \leq 2

Piecewise function:

The piecewise function is:

C(x) = C(x) = \left \{ {{4, 0< x \leq 1} \atop {4 + 8x, 1 < x \leq 2}}\right.

7 0
2 years ago
How would i solve a problem like this?
SVETLANKA909090 [29]

The number of combinations of size k that you can make with n items is given by the so-called binomial coefficient,

\dbinom nk = \dfrac{n!}{k!(n-k)!}

• n! is the number of ways of permuting n items.

• (n-k)! is the number of ways of permuting all but k of the n items.

Dividing n! by (n-k)! then gives the number of ways of permuting only k of the total n items.

• k! is the number of ways of permuting k items.

Dividing \frac{n!}{(n-k)!} by k! then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20

ways of selecting any 3 girls from the total 6 girls in the entire group of people.

There are

\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21

was of selecting any 2 boys from the total 7 boys.

Then there are

\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287

Then the probability of selecting such a committee at random is

\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263

8 0
1 year ago
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