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7nadin3 [17]
3 years ago
10

Question is in the pictureWILL GIVE BRAINLIST OR HOWEVER ITS SPELT

Mathematics
2 answers:
NARA [144]3 years ago
8 0
The first one is the answer
DanielleElmas [232]3 years ago
3 0

Answer:

B

Step-by-step explanation:

it’s either B or C this is a weird one. If it’s was something like 7^ 3/4 there would be a 4 outside the square root and the main number inside  with the little 3 next to the 7.

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The arena wants to estimate the previous two month’s ticket-sale income. in total, 71,115 people attended different events at th
OleMash [197]
There was a total of 71,115 people who attended the different  events in the arena for the past two months,

The estimated average ticket price is $ 21.15.

We need to solve for the estimated arena's ticket sale income and the solution is shown below:
Ticket Sale Income = 71, 115 * $21.15
Ticket Sale Income = $1,504,082
3 0
3 years ago
Given that XV = 11 and XQ = 3, is X the incenter or the circumcenter of /? Find XR. A. circumcenter; XR = 3 B. incenter; XR = 3
guajiro [1.7K]

Answer:

The answer is C.

I did the practice.

7 0
2 years ago
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
2 years ago
What are the solutions to the equation (x – 6)(x + 8) = 0?
Ghella [55]

Answer:

x₁ = 6 and x₂ = 8

Step-by-step explanation:

See the picture below

8 0
3 years ago
S= 4LW + 2WH; S=152, L=8, W=4. Find H
Katyanochek1 [597]
H is equal to 3. need the solution?
5 0
2 years ago
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