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jenyasd209 [6]
3 years ago
9

1 A high school talent show lasted a total of t minutes. There were 2 intermissions that were each the length of 8 the talent sh

ow. Which function can be used to find the number of minutes the talent show lasted, not including intermissions?​
Mathematics
1 answer:
Sophie [7]3 years ago
4 0

Answer:

I think f(t)=1/4t

Step-by-step explanation:

Because of you multiply 1/8 times 2, it gives you 1/4. Not 100% sure tho

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Equal to a+b on the algebra
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What is the solution to the system of equations? HELP
notsponge [240]
I will solve your system by substitution.<span><span>x=<span>−2</span></span>;<span>y=<span><span><span>23</span>x</span>+3</span></span></span>Step: Solve<span>x=<span>−2</span></span>for x:Step: Substitute<span>−2</span>forxin<span><span>y=<span><span><span>23</span>x</span>+3</span></span>:</span><span>y=<span><span><span>23</span>x</span>+3</span></span><span>y=<span><span><span>23</span><span>(<span>−2</span>)</span></span>+3</span></span><span>y=<span>53</span></span>(Simplify both sides of the equation)

Answer:<span><span>x=<span>−<span><span>2<span> and </span></span>y</span></span></span>=<span>5/3
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so the answer is B (the second choice)
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4 0
3 years ago
Read 2 more answers
a train leaves little rock, arkansas and travels north at 70 kilometers per hour. another train leaves at the same time and trav
jekas [21]
45 hours.
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8 0
4 years ago
Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we
Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N(\mu = 7 lbs , \sigma^{2}  = 0.875^{2}  lbs)

The standard normal z distribution is given by;

              Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, X bar = sample mean weight

             n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{7.5-7}{\frac{0.875}{\sqrt{4} } }  ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0
4 years ago
Two fair cubes, numbered 1 through 6, are tossed at the same time in order to advance on a game board. What is the probability t
DaniilM [7]
If the cubes are rolled independently, and assuming any face on the cube is as likely to occur as any other, then there is a 1/6 probability of rolling a 6, and rolling a 6 on both simultaneously is 1/6 * 1/6 = 1/36.
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3 years ago
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