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Bas_tet [7]
3 years ago
9

Solve. 4y 16 > –8 a. y > –6 b. y > –2 c. y > 2 d. y > 6

Mathematics
1 answer:
joja [24]3 years ago
7 0
Sorry if no one helped u well im here lets get started
I would go with C.y>2
26>-8 true
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PLEASE HELP! giving BRAINLIEST AND GIVE CORRECT ANSWER AND EXPLAIN!
11111nata11111 [884]

The exercise is basiacally asking for the perimeter of the garden. The perimeter of a rectangle is twice the sum of its dimensions, so we have

P=2(6+11)=2\cdot 17=34

7 0
3 years ago
Of the employees who work at stalling printing 90% attened the safety procedures meeting. If 63 employees attended the meeting h
My name is Ann [436]
90% of what is 63....

0.90x = 63
x = 63 / 0.90
x = 70....so 70 ppl work at stalling printing
6 0
2 years ago
If you flipped the graph y=x^2+2x-2 vertically, you would get the graph y=-(x^2 +2x-2). true or false
Mars2501 [29]
If you flipped the graph y=x^2+2x-2 vertically, you would get the graph y=-(x^2+2x-2) this is True. 
5 0
2 years ago
A rectangular metal plate is measured to be 7.6cm long and 3.1cm wide, both correct to one decimal place.
ElenaW [278]

Answer:

We know that the rectangular plate has measures of:

length = 7.6 ± 0.05 cm

width = 3.1 ± 0.05 cm

(the error is 0.05cm because we know that both measures are correct to one decimal place)

First, the upper bound of the length is equal to the measure of the length plus the error, this is:

L = 7.6 cm + 0.05 cm = 7.65 cm

The upper bound of the area is the area calculated when we use the upper bound of the length and the upper bound of the widht.

Remember that the area for a rectangle of length L and width W, is:

A = W*L

Then the upper bound of the area is:

A = (7.6cm + 0.05cm)*(3.1cm + 0.05cm) = 10.8 cm^2

5 0
3 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
3 years ago
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