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atroni [7]
3 years ago
5

Find the experimental probability of rolling a 4

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0
<span />the probability of rolling a 4 = 5 times.
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14 - 8i

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Which of the following inequalities matches the graph?
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c

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A missionary used his plane to take an injured girl 855 kilometers to the hospital. If he flew at an average speed of 171 kilome
melisa1 [442]

Step-by-step explanation:

171 km/h

travel distance = 855 km

855 / 171 = 5 hours

it took him 5 hours to fly the 855 km (theoretically to the hospital).

normally, with a plane, you can't land on or at a hospital. you would need a helicopter to be able to do that.

so, he could only fly to a nearby airport, and then travel from there to the hospital. but we have no information about that.

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2 years ago
Can someone help me with this??
dusya [7]

Answer:

Carlos and Pamela drove 120 miles on the first day, 240 miles on the second day, and 290 miles on the third day.

Step-by-step explanation:

Let x be the number of miles driven on the first day.

Then they drove twice as many miles, or 2x on the second day

and 50 miles more than the second day's, so 2x + 50

The total is 650 across all three days, so we'll take the sum.

x + 2x + (2x + 50) = 650

Combine like terms on the left

5x + 50 = 650

Subtract 50 on both sides

5x = 600

Divide by 5 on both sides

x = 120

Check work:

120 + 2(120) + 2(120) + 50 = 650

120 + 240 + 240 + 50 = 650

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So they drove 120 miles on the first day

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4 0
3 years ago
Two positive integers have a sum of 10 and a product of 21. What are the integers?
lara [203]

Answer:

<h2>3 and 7</h2>

Step-by-step explanation:

x,y-\text{positive integers}\\\\\text{The system of equations:}\\\\\left\{\begin{array}{ccc}x+y=10&\text{subtract}\ y\ \text{from both sides}\\xy=21\end{array}\right\\\\\left\{\begin{array}{ccc}x=10-y&(1)\\xy=21&(2)\end{array}\right\\\\\text{substitute (1) to (2):}\\\\(10-y)(y)=21\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\\\10y-y^2=21\qquad\text{subtract 21 from both sides}\\\\-y^2+10y-21=0\qquad\text{change the signs}\\\\y^2-10y+21=0\\\\y^2-3y-7y+21=0\\\\y(y-3)-7(y-3)=0

(y-3)(y-7)=0\iff y-3=0\ \vee\ y-7=0\\\\y-3=0\qquad\text{add 3 to both sides}\\\\y=3\\\\y-7=0\qquad\text{add 7 to both sides}\\\\y=7\\\\\text{Put the value of}\ y\ \text{to (1):}\\\\x=10-3=7\ or\ x=10-7=3

7 0
3 years ago
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