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dexar [7]
4 years ago
11

If 10 miles is about 16 km and the distance between two towns is 45 miles use a ratio table to find the distance between the tow

ns in kilometers
Mathematics
1 answer:
guapka [62]4 years ago
7 0
The ratio of m to km is
10/16
5:8
So for every 5 miles there are 8 km

45/5 = 9
9 * 8 = 72

Let's Check the answer:

45/72
5:8

So, its correct. The answer is 72 km.
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Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
krok68 [10]

Answer:

(x, y, z) =(1, -1,2)

Step-by-step explanation:

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8 0
3 years ago
Four and two thirds add one and three fourths
Allushta [10]

Answer:

4 2/3 + 1 3/4 = 6 5/12

3 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Alenkasestr [34]

Answer:

A = -10

Step-by-step explanation:

x^2 + 10x + 24 = 0

To find the roots, we need to factor the equation.

What 2 numbers multiply to 24 and add to 10

6*4 = 24

6+4 = 10

(x+6) (x+4)=0

Using the zero product property

x+6 = 0  x+4 = 0

x = -6   x = -4

We want the sum when the roots are added

-6+ -4 = -10

6 0
3 years ago
-6+14 =12-8x what is x
navik [9.2K]
8=12-8x
-4=8x
-4÷8=8x÷8
-0.5=x
5 0
3 years ago
Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random
lesya692 [45]
Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes P_B(A).

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
P(A|B)= \frac{P(A \cap B)}{P(B)}

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is \frac{1}{3}

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

Therefore, P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}


Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}

</span><span>P(D)= \frac{1}{3}</span><span>

Therefore, P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}</span>
3 0
4 years ago
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