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artcher [175]
3 years ago
10

Please Help ITS DUE TOMORROW!!!

Mathematics
1 answer:
Kaylis [27]3 years ago
4 0
The answer is on the paper

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Does this have a solution? 4(x+3)+2x=6(x+2)
Tamiku [17]

Answer:

11x

Step-by-step explanation:

4(x+3)+2x=6(x+2)

BIDMAS

x+3=3x

x+2=2x

3x+2x+2x7x

7x+4=11x

i think im so sorry if this isnt helpful

6 0
4 years ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and
charle [14.2K]

Answer:

A) 0.5737

B) 0.9884

Step-by-step explanation:

We are given that an engineer is going to redesign an ejection seat for an airplane.  The new population of pilots has normally distributed weights with a mean of 160 lb and a standard deviation of 27.5 lb i.e.;                                                 \mu = 160 lb  and \sigma = 27.5 lb

(A) We know that Z = \frac{X - \mu}{\sigma} ~ N(0,1)

Let X = randomly selected pilot  

If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)

P(150 < X < 201) = P(X < 201) - P(X <= 150)

P(X < 201) = P( \frac{X - \mu}{\sigma} < \frac{201 - 160}{27.5} ) = P(Z < 1.49) = 0.9319

P(X <= 150) = P( \frac{X - \mu}{\sigma} < \frac{150 - 160}{27.5} ) = P(Z < -0.3636) = P(Z > 0.3636) = 0.3582

Therefore, P(150 < X < 201) = 0.9319 - 0.3582 = 0.5737 .

(B) We know that for sampling mean distribution;

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

If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);

 P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)

  P(X bar < 201) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{201 - 160}{\frac{27.5}{\sqrt{39} } } ) = P(Z < 9.311) = 1 - P(Z >= 9.311)

                                                                                    = 0.999995

P(X bar <= 150) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{150 - 160}{\frac{27.5}{\sqrt{39} } } ) = P(Z < -2.2709) = P(Z > 2.2709)

                                                                                          = 0.0116

Therefore,  P(150 < X bar < 210) = 0.999995 - 0.0116 = 0.9884

C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.

8 0
3 years ago
. Center at (-4,-1), radius = 2​
valkas [14]

Consider we need to find the equation of the circle.

Given:

The Center at (-4,-1), radius = 2​.

To find:

The equation of the circle.

Solution:

The standard form of a circle is

(x-h)^2+(y-k)^2=r^2

Where, (h,k) is the center of the circle and r is the radius of the circle.

Putting h=-4, k=-1 and r=2, we get

(x-(-4))^2+(y-(-1))^2=(2)^2

(x+4)^2+(y+1)^2=4

Therefore, the equation of the circle is (x+4)^2+(y+1)^2=4.

6 0
3 years ago
What is the slope-intercept form (y= mx+ b) of the equation of the line? Hint: Find the y-intercept. Then count up and over (y/x
Anna35 [415]

Answer:

12

Step-by-step explanation:

didnt do it

8 0
3 years ago
Can anyone help me with this please
zimovet [89]
Let’s see, so x=6

The equations should be written out as
2(10) = 8(4) - 2(6)

The x is replaced by 6, since in this problem x equals 6, and thenceforth are the same. When solved:

20 = 32 - 12

Which means:

20 = 20

Which is correct.
5 0
3 years ago
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