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Slav-nsk [51]
2 years ago
5

Hi can someone help me find domain and range

Mathematics
1 answer:
dezoksy [38]2 years ago
8 0

Answer:

Step-by-step explanation:

Domain is ( - ∞ , ∞ )

Range is [ 0 , 1 ] ∪ ( 2 , ∞ )

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0.61566147 Round to the nearest ten-thousandth. I JUST NOTICED IM ON THE SCOREBOARD!!!!!
Pie

Answer:

0.6157

Step-by-step explanation:

3 0
3 years ago
What is -5 3/4 - 3 1/2​
lubasha [3.4K]

Simplify the expression.

Exact Form:

−37/4

Decimal Form:

−9.25

Mixed Number Form:

−9 1/4

3 0
4 years ago
Read 2 more answers
HEY LOL PLEASE ANSWER THESE. PLEASE IM BEGGING YOU. HELP ME
Alexandra [31]

Answer:

a. ∆ABC ~ ∆A'B'C'

b. Enlargement

c. Scale factor = 4

Step-by-step explanation:

a. <A corresponds to <A'

<B corresponds to <B'

<C corresponds to <C',

Therefore, the similarity statement would be: ∆ABC ~ ∆A'B'C'

b. The image of the dilation is an enlargement of the original figure because the side lengths of ∆A'B'C are larger than the corresponding side lengths of ∆ABC.

c. Scale factor = the ratio of the corresponding sides

Thus:

Scale factor = A'B'/AB = B'C'/BC = A'C'/AC

Scale factor = 24/6 = 48/12 = 60/15 = 4

Scale factor = 4

5 0
3 years ago
Jana And Jordan had 23 oranges Jana had 7 fewer oranges than Jordan how many oranges did each girl have
Solnce55 [7]

Answer:

Jana had 8 oranges, and Jordan had 15 oranges.

Step-by-step explanation:

First, identify what you know:

1) In total, Jana and Jordan had 23 oranges.

2) Jana had 7 less oranges than Jordan.

We can create a formula, where x equals the number of oranges Jana has.

23 = x + (x + 7)

16 = x + x (subtracted 7 from both sides)

x = 8 (divided both sides by 2)

So, now we know Jana had 8 oranges, which is 7 less than Jordan.

8 + 7 = ?

? = 15

Jana had 8 oranges, while Jordan had 15, for a total of 23 oranges.

3 0
3 years ago
Suppose p(a) = 0.40 and p(a 
sveticcg [70]
Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

P(A\cup B)+P(A\cap B)=P(A)+P(B)


For parts (a) and (b), you're given everything you need to determine P(B).

For part (c), if A and B are mutually exclusive, then P(A\cap B)=0, so P(A\cup B)=P(A)+P(B). If the given probability is P(A\cup B)=0.55, then you can find P(B)=0.15. But if this given probability is for the intersection, finding P(B) is impossible.


For part (d), if A and B are independent, then P(A\cap B)=P(A)\cdot P(B).
8 0
3 years ago
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