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Alborosie
3 years ago
5

Please help I am confused

Mathematics
1 answer:
anzhelika [568]3 years ago
7 0
X=11
So do I think hope this helps!!
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Worth 10 need help due today
cricket20 [7]
E. all of the points above
4 0
3 years ago
Can someone help me out
Harrizon [31]

Answer:

82.8

Step-by-step explanation:

area of the parallelogram = base × height

= 13.8×6

= 82.8 in²

Answered by GAUTHMATH

8 0
3 years ago
i need help with this equation please there are two more possible answers that were cut off they are 17,2% and 19,5%
Margarita [4]

Consider that the experimental probability of an event is based upon the previous trials and observations of the experiment.

The experimental probability of occurrence of an event is given by,

\text{Probability of an event}=\frac{\text{ Number of outcomes that favoured the event}}{\text{ Total number of trials or outcomes}}

As per the problem, there are a total of 1230 trials of rolling a dice.

And the favourable event is getting a 2.

The corresponding experimental probability is calculated as,

\begin{gathered} P(\text{ getting a 2})=\frac{\text{ No. of times 2 occurred}}{\text{ Total no. of times the dice is thrown}} \\ P(\text{ getting a 2})=\frac{172}{1230} \\ P(\text{ getting a 2})\approx0.13984 \\ P(\text{ getting a 2})\approx13.98\text{ percent} \end{gathered}

Thus, the required probability is 13.98% approximately.

Theref

7 0
1 year ago
50 POINTS! Plot function h on the graph.<br><br> NO LINKS
tangare [24]

See attachment for the graph of the piecewise function h(x)

<h3>How to plot the function?</h3>

The function is given as:

h(x) = | -4,    x < 3

         | x + 5,    x >= 3

The above function is a piecewise function.

It has 2 separate functions at two domains

This means that we plot the sub-functions in the piecewise function at their respective domain

See attachment for the graph of the piecewise function h(x)

Read more about piecewise function at:

brainly.com/question/27262465

#SPJ1

3 0
2 years ago
Hello! Could someone help me out with part d? Please show your work!
koban [17]

Answer:

1

Step-by-step explanation:

Part d

∫f′(t)dt   as long as the function is continuous

Let u = 6-2x

du = -2dx  so -1/2 du = dx

upper limit = 6- 2(4) = 6-8 = -2

lower limit = 6 - 2(2) = 6 - 4 = 2

∫f′(u)(-1/2)du  limits -2  to 2  =-1/2( f(-2) - f(2))

f(-2) is 1

f(2)  is 3

-1/2(f(-2) - f(2)) = -1/2( 1-3) = -1/2(-2) = 1

4 0
3 years ago
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