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just olya [345]
2 years ago
9

Forty percent of prison inmates were unemployed when they entered prison. If 5 inmates are randomly selected. Find these possibi

lities.
a. Exactly 3 were unemployed.
b. At most 4 were unemployed.
c. At least 3 were unemployed.
d. Fewer than 2 were unemployed.
Mathematics
2 answers:
melamori03 [73]2 years ago
8 0
D. Fewer than 1 were unemployed
irinina [24]2 years ago
6 0
D, I think because 40 % of 100 is 4 then you divide them into in half and get every 2/5 inmates were unemployed before they entered prison
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Candy draws a square design with a side length of x inches for the window at the pet shop. She takes the design to the printer a
Inga [223]

Answer:

This is late, but for anyone else that needs it, it's B. 4x-5

8 0
2 years ago
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Orin solved an equation and justified his steps as shown in the table.
sergij07 [2.7K]

Answer:

Step-by-step explanation:

Given the equation as

\frac{3x}{5} -3=12

apply multiplication property of equality where you multiply every term by 5

\frac{3x}{5}*5 -3*5=12*5

3x-15=60------------------apply addition property of equality

3x-15+15=60+15

3x=75--------------------------appy division property of equality by dividing both sides by 3

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x=25

6 0
2 years ago
Identify the independent and dependent variables in the following hypothesis: "People who exercise regularly are more likely to
JulijaS [17]

Answer:

Independent

dependent

Step-by-step explanation:

The scenario illustrates that the weight for people who exercises regularly tends is normal or below normal. It means that the weight of a person depends on how much a person exercises i.e. if person exercises more the weight is reduced and if not then weight will be gained. Thus, exercise is the independent variable while weight is dependent variable.

7 0
2 years ago
Given that tangent theta = negative 1, what is the value of secant theta, for StartFraction 3 pi Over 2 EndFraction less-than th
Rashid [163]

Answer:

sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}

Step-by-step explanation:

Start by noticing that the angle \theta is on the 4th quadrant (between \frac{3\pi}{2} and 2\pi. Recall then that in this quadrant the functions tangent and cosine are positive, while the function sine is negative in value. This is important to remember given the fact that tangent of an angle is defined as the quotient of the sine function at that angle divided by the cosine of the same angle:

tan(\theta)=\frac{sin(\theta)}{cos(\theta)}

Now, let's use the information that the tangent of the angle in question equals "-1", and understand what that angle could be:

tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\-1=\frac{sin(\theta)}{cos(\theta)}\\-cos(\theta)=sin(\theta)

The particular special angle that satisfies this (the magnitude of sine and cosine the same) in the 4th quadrant, is the angle \frac{7\pi}{4}

which renders for the cosine function the value \frac{\sqrt{2} }{2}.

Now, since we are asked to find the value of the secant of this angle, we need to remember the expression for the secant function in terms of other trig functions: sec(\theta)=\frac{1}{cos(\theta)}

Therefore the value of the secant of this angle would be the reciprocal of the cosine of the angle, that is: sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}

6 0
2 years ago
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Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,
grandymaker [24]

Solution:

There is no saddle point (DNE). However, there is local maximum at (1, 1/2) for the given function.

Explanation:

we have function of two variables f(x,y)= 9-2x+4y-x^2-4y^2

we will find the values by partial derivative with respect to x,y,xy

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to find the saddle point we should first find the critical points so equate

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we get x= 1  and y =1/2 so, critical points are (1,1/2)

to find local maximum or minimum we have to find f_{xx},  f_{yy} and f_{xy}

formula is f_{xx} *f_{yy} - f^{2_{xy} } =0

f_{xx} = -2

f_{yy} = -8

f^{2_{xy} } =0

putting values in formula

(-2)*(-8) -0 =16 > 0, and f_{xx}< 0  and f_{yy}<0

so, here we have local maximum

we have no saddle point for this function by using the same formula we used to find extrema.



4 0
2 years ago
Read 2 more answers
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