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uysha [10]
3 years ago
9

Dont understandd helppp​

Mathematics
1 answer:
Romashka [77]3 years ago
8 0

Answer:

y\leq -12 or y\geq -4

  1. Subtract 6 from both sides 3|y+8|+6-6\geq18-6
  2. Simplify 3|y+8|\geq12
  3. Divide both sides by 3
  4. Simplify |y+8|\geq4
  5. Apply absolute rule y+8\leq -4 or y+8\geq4
  6. Combined the interval y\leq -12 or y\geq -4

<h2><em>y</em>\leq<em> -12 or y</em>\geq<em> -4</em></h2><h2><em>:) Hope You Enjoy :)</em></h2><h2> <em>They are both real numbers :)</em></h2>
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Answer:

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4 years ago
Please answer this it would be very helpful if you did.
Dmitriy789 [7]

Answer:

See below ↓↓↓

Step-by-step explanation:

<u>Probability of getting heads</u>

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<u>Probability of getting an odd number</u>

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5 0
2 years ago
The weights of adult male beagles are normally distributed, with a mean of 25 pounds and a standard deviation of 4 pounds.
Andrews [41]

Answer:

A) 8

Step-by-step explanation:

Mean weight = u = 25 pounds

Standard Deviation = \sigma = 4 pounds

We have to find how many beagles out of 75 will weigh more than 30. Since the data is normally distributed, we can use z score to find this value. First we will what is the percentage(probability) of a randomly selected beagle to weigh more than x = 30 pounds, using this percentage we can then find our answer.

The formula for z score is:

z=\frac{x-u}{\sigma}

Using the values, we get:

z=\frac{30-25}{4}=1.25

So, P(Weight > 30) is equivalent to P(z > 1.25). Using the z table, we can write:

P(z > 1.25) = P(Weight > 30) = 0.1056

This, 0.1056 or 10.56% of the beagles are expected to weigh more than 30 pounds.

So,out of 75 beagles, 10.56% of 75 are expected to weigh more than 30 pounds.

10.56% of 75 = 0.1056 x 75 = 7.92 = 8 (rounding to nearest integer)

Therefore, out of 75 beagles 8 are expected to weigh more than 30 pounds.

6 0
3 years ago
Find the average rate of change of the function over the given interval.. h(t) = cot t, intervals given [pi/4, (3pi)/4]. I've go
Phantasy [73]
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3 0
4 years ago
A man can walk from A to B and back in a certain time at 7 km/hr.If he walks from A to B at 8 km/hr and returns from B to A at 6
DerKrebs [107]
We calculate the speed by dividing the distance over time:
s = d/t
So the distance described in the problem is always the same, A to B and B to A.
But we are told that;
7 = d/t
7 = 2d/(t + 2)
that is, the first equation say that at speed 7 km/h a distance d is walked in a time t
the second equation say that at a average speed of 7 (that is 8 on one way and 6 in the other: 8 + 6 = 14, half of it), twice the distance is walked in a time equal to the first time plus 2 minutes.
So we have a system of linear equations, 2 of them with two unknowns, we can solve that:
7 = d/t
7 = 2d/(t + 2<span>)
</span>lets simplify them:
7t = d
7(t + 2) = 2d
7t = 2d - 14
we substitute the first in the second:
<span>7t = 2d - 14
</span><span>7t = d
</span>so:
d = 2d - 14
d = 14
so the distance between A and B is 14 km
4 0
4 years ago
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