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Alex787 [66]
3 years ago
8

I need help please!!!!!!

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
8 0

Answer:

THE ABSWER IS OPTIOM 1 AINCE ITS BEE SIMPLE AND IR CAN BE MORE IF U CAN TRY BUT U DONT NEED TK

Tamiku [17]3 years ago
6 0

Answer:

Step-by-step explanation:

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A wall is 4 1/2 feet long and has an area of 15 3/4 square feet. What is its height in feet?
IRINA_888 [86]

Answer:

3 1/2

Step-by-step explanation:

if you multiply 4 and 3 you 12 if you multiply 1/2 times 3 you get 1 1/2 and 1 1/2 + 12 = 13 1/2 and then you have to find a fraction that when multiplied by 4 1/2 you get 2 1/4 and 4 1/2 × 1/2 = 2 1/4 so 2 1/4 + 13 1/2 = 15 3/4

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3 years ago
Which is true statement about 7 and 14
STatiana [176]
They are both a divisible by 7
8 0
4 years ago
10, 7, 3.5<br>what's the answer ​
Alona [7]

Answer:

7

Step-by-step explanation:

3 0
3 years ago
A veterinarian’s assistant made a table that shows the animals seen in the office in one week. What is the probability that the
horsena [70]

Answer:

The probability that the pet seen was sick is 53.57%.

Step-by-step explanation:

The probability of an event <em>E</em> is the ration of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

Here,

n (E) = number of favorable outcomes

N = total number of outcomes

Denote the events as follows:

<em>C</em> = the pet is a cat

<em>D</em> = the pet is a dog

<em>O </em>=<em> </em>the pet is some other animal

<em>S</em> = the pet is sick.

The data provided is summarized as follows:

n (C) = 28

n (C ∩ S) = 3

n (D) = 42

n (B ∩ S) = 4

n (O) = 24

n (O ∩ S) = 8

Compute the probability that a cat was sick as follows:

P(C\cap S)=\frac{n(C\cap S)}{n(C)}=\frac{3}{28}

Compute the probability that a dog was sick as follows:

P(D\cap S)=\frac{n(D\cap S)}{n(D)}=\frac{4}{42}=\frac{2}{21}

Compute the probability that another animal was sick as follows:

P(O\cap S)=\frac{n(O\cap S)}{n(O)}=\frac{8}{24}=\frac{1}{3}

Compute the probability that the pet seen was sick as follows:

P (S) = P (C ∩ S) + P (D ∩ S) + P (O ∩ S)

       =\frac{3}{28}+\frac{2}{21}+\frac{1}{3}\\=\frac{9+8+28}{84}\\=\frac{45}{84}\\=0.5357

Thus, the probability that the pet seen was sick is 53.57%.

8 0
4 years ago
Read 2 more answers
What is the median of this set of data? 1, 2, 5, 6, 9 1 5 8 9
gavmur [86]

Answer:

5.5

Step-by-step explanation:

1, 2, 5, 5, 6, 8, 9, 9

5+6=11

11 divided by 2= 5.5

4 0
4 years ago
Read 2 more answers
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