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Makovka662 [10]
2 years ago
9

Which of the following values of d will result in a true statement when substituted into the given equation?

Mathematics
2 answers:
saw5 [17]2 years ago
8 0
I think it's D cause the parentheses always comes first so 2-d)=2 hope that helps
DIA [1.3K]2 years ago
7 0
The answer is C) d = 2
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The cafeteria can serve 32 lunches in 5 minutes. How long will it take them to serve 57 lunches?
PolarNik [594]

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Around 364

Step-by-step explanation:

So to fund the answer, you need to find how much they can eat per minute: 32/5 = 6.4 (around 6 lunches)

Then you multiply this by 57: 57x6.4 = 364.8

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There are 10 less red skittles than orange skittles in the bag. The orange skittles are also twice the number of red skittles. F
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3 years ago
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GIVING BRAINLIEST TO THE PERSON THAT CAN EXPLAIN THIS THE BEST (:
Amanda [17]

Answer:

25π-24

Step-by-step explanation:

from the figure,

radius of the circle(r)=diameter of the circle(d)/2

=10/2

=5

Area of the circle(A)=πr^2

=π5^2

=25π

Again,

base of the right angle triangle(b)=6

perpendicular of right angle triangle(p)=8

we also know,

the area of right angle triangle = 1/2*base*height

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5 0
2 years ago
Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i
vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable <em>X</em> be defined as the number of students at a particular college who are smokers

The random variable <em>X</em> follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10

Thus, a Normal approximation to binomial can be applied.

So,  

X\sim N(\mu=600, \sigma=\sqrt{120})

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   =P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

8 0
2 years ago
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