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Gelneren [198K]
3 years ago
14

The National Safety Council of Ireland found that young men were responsible in 57% of accidents they were involved in. The NSC

Web site made this claim: "Young men are responsible for over half of all road accidents." From the selection below, what can you conclude from the NSC statement
Mathematics
1 answer:
lakkis [162]3 years ago
7 0
Males are more reckless than females.
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Find the area of each sector.
nevsk [136]

Answer:

225

Step-by-step explanation:

360 - 135 = 225

7 0
3 years ago
U.S. Customary unit conversion with mixed number values: One... A cook needs 7 cups of vegetable stock for a soup recipe. How mu
lapo4ka [179]

The pints of vegetable stock the chef needs is 3\frac{1}{2}.

<h3>How many pints does he need?</h3>

The first step is to determine the unit of conversion of cups to pints.

1 cup = 0.5 pints

To convert cup to pints, multiply 7 by 0.7

7 x 0.5 = 7/2 = 3\frac{1}{2}

To learn how to convert units, please check: brainly.com/question/25993533

#SPJ1

4 0
2 years ago
I don’t get it (show work)!!
Mashcka [7]

Answer:

8r+8t

Step-by-step explanation:

Add Like Terms

Hope This Helps!  Have A Nice Day!!

7 0
3 years ago
Read 2 more answers
Find the quotient. <br> -300x 5 y 4 ÷ xy 2 <br> -150x6y6<br> -300x6y6<br> 300x4y2<br> -300x4y2
Helen [10]
\frac{-300x^{5}y^{4}}{xy^{2}} = -300x^{4}y^{2}

The answer is D.
4 0
4 years ago
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A 10.0 kg and a 2.0 kg cart approach each other on a horizontal frictionless air track. Their
Flura [38]

the  final speed in m/s of the 10.0 kg is 2.53 m/s .

<u>Step-by-step explanation:</u>

Here we have , A 10.0 kg and a 2.0 kg cart approach each other on a horizontal friction less air track. Their  total kinetic energy before collision is 96 ). Assume their collision is elastic. We need to find What is the  final speed in m/s of the 10.0 kg mass if that of the 2.0 kg mass is 8.0 m/s . Let's find out:

We know that in an elastic collision :

⇒ Total kinetic energy before collision  = Total kinetic energy after collision

⇒ 96 = \frac{1}{2}M_1(v_1)^2 + \frac{1}{2}M_2(v_2)^2

⇒ 96 = \frac{1}{2}(10)(v_1)^2 + \frac{1}{2}(2)(8)^2

⇒ 96=5(v_1)^2 + 64

⇒ 5(v_1)^2  =32

⇒ (v_1)^2  =6.4

⇒ v_1  =2.53 m/s

Therefore , the  final speed in m/s of the 10.0 kg is 2.53 m/s .

8 0
3 years ago
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