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ivolga24 [154]
3 years ago
11

in Mr.Kleins class, 40% of the students are boys. What decimal represent the portion of the students that are girls

Mathematics
1 answer:
mezya [45]3 years ago
6 0

Answer: 0.6

Step-by-step explanation: Since 40 percent are boys which is .4 in decimal form that would mean there are 60 percent of girls left which is .6 in decimal form.

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Value of x is less than 14
nalin [4]

Hey there!

Value of x is less than 14 would look like this:

x < 14

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lisov135 [29]
The answer is C. 91.46 !!!
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What values for PQ and RL make the following equation true px to the power of rtwo+wx+r equals(3x+1)(3x+2)HURRY
zhannawk [14.2K]

P= 3

Q= -14

R= -5

Step-by-step explanation:

Just solve the equation on the right side, you'll get:

=3x^2 -15x +x -5

=3x^2 -14x -5

Now compare it to the equation on the left side,

The coefficient of x^2 on the left is "p", on the right that we just solved is 3.

Same for "q" which is the coefficient on the left, on the right it's -14.

And for "r" it's -5 from the equation on the right.

6 0
3 years ago
In order to determine whether or not a variable type of process is in control (whereoutput can be measured, not just classified
Tema [17]

Answer:

i dont know

Step-by-step explanation:

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6 0
4 years ago
Suppose that the lifetime of tv tubes are normally distributed with a standard deviation of 1.1 years. Suppose that exactly 20%
IrinaK [193]

Answer:

4.924 years

Step-by-step explanation:

Lets denote X the lifetime of a tv tube (In years). X has distribution N(\mu, 1.1) , with \mu unknown.

We know that P(X < 4) = 0.2. Using this data, we can find the value of \mu throught standarization.

Lets call Z = \frac{X - \mu}{1.1} the standarization of X. Z has distribution N(0,1), and its cummulative function, \phi is tabulated. The values of \phi can be found in the attached file.

P(X < 4) = P(\frac{X - \mu}{1.1} < \frac{4 - \mu}{1.1}) = P(Z < \frac{4 - \mu}{1.1}) = \phi(\frac{4 - \mu}{1.1}) = 0.2

The value q such that \phi (q) = 0.2 doesnt appear on the table. We can find it by using the symmetry of the normal density function. The opposite of q, -q must verify that \phi(-q) = 0.8 , hence -q must be equal to 0.84. Thus, q = -0.84

But this value of q should match with the number \frac{4 - \mu}{1.1} , so we have

\frac{4- \mu}{1.1} = -0.84

4 - \mu = 1.1 * (-0.84) = -0.924

\mu = 4 - (-0.924) = 4.924

Thus, the expected lifetime of TV tubes is 4.924 years.

I hope this works for you!

Download pdf
7 0
4 years ago
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