Inequalities

1 answer:

4 0

for all of these, you just have to divide by -1 to make the variable positive however you have flip the inequality sign when you divide by a negative number.

c > 1

-2 > v

y > 1

z > 6

1 > w

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I do not even know where to begin. Please help.

xenn [34]

F(x) = 2x + 1 when x <= -5
Now -10 is less than -5 so we plug x = -10 into 2x +1:-

f(-10) = 2(-10) + 1 = -19 answer

Now f(x) = x^2 when -5< x < 5
x =2 is within this range so f(2) = 2^2 = 4 answer

You can calculate the other 3 values by a similar method

5 0

2 years ago

.................help

Ann [662]

Answer:

Step-by-step explanation:

He found out how many people who play 2+ instruments are 8th graders rather than how many 8th graders play 2+ instruments.

8 0

3 years ago

A stack of books is 5 feet high of the average book is 3 inches Hugh how many books are in the stack

allochka39001 [22]

Answer:

Step-by-step explanation:

1) 5 feet is 60 inches

2)Divide 60 inches (total inches in one stack) by the 3 inches ( inches in the Average book)

3) 60/3=<u>20</u>

7 0

3 years ago

Read 2 more answers

A minivan can hold up it 7 people. How much minivan would it take to fit 45 people

iVinArrow [24]

Answer:

6.42857143 so you can just say 7

Step-by-step explanation:

45/7 = 6.42857143

3 0

3 years ago

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec

Sphinxa [80]

<span>1.

P(</span>at least one of the calculators is defective)=

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

=n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways,

where $C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}= \frac{35*34*33*32}{4!}= \frac{35*34*33*32}{4*3*2*1}= 52,360$

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

$C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825$

4. so, P(none of the selected calculators is defective)= $\frac{52,360}{292,825} =0.18$

5. P(at least one of the calculators is defective)=

1- P(none of the selected calculators is defective)=1-0.18=0.82

Answer:0.82

7 0

3 years ago