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svetlana [45]
3 years ago
11

Based on the housing data below, which equation can be used to calculate

Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0

Answer: y= 0.074x + 50.48

Step-by-step explanation:

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Find the value of x.<br> 3<br> x = [?]
satela [25.4K]

Answer:

12

Step-by-step explanation:

According to Euclidian theorem h^2 = 3*x (h : height)

36 = 3x divide both sides by 3

12 = x

4 0
3 years ago
The product of 6 and seven times c
Sunny_sXe [5.5K]

Answer:

42

Step-by-step explanation:

8 0
4 years ago
I'm trying to help my friend
diamong [38]
<span><span>4<span>a<span><span>​^</span><span>​​</span></span></span><span>b<span><span>​<span><span>​3</span>​<span>​^7</span><span>​</span></span></span><span>​​</span></span></span><span>c<span>​^3</span></span></span><span><span>​​</span></span></span>
4 0
3 years ago
Adrian dog ate 40 cans of dog food in 30 days how many cans of dog food for 6 days
babunello [35]
Alright! In order to solve this problem you first have to find how how many cans of dog food he ate in one day. To do that you simply divide 40 cans by 30 days:

\frac{40}{30} → \frac{4}{3}

If the dog ate \frac{4}{3} per day, then to find out how many cans he ate in 6 days to multiply \frac{4}{3} by 6:

\frac{4}{3} x 6 → \frac{4}{3} x \frac{6}{1} = \frac{24}{3}

\frac{24}{3} simplified is 8!

Adrian's dog at 8 cans of food in 6 days.
5 0
3 years ago
You roll a six-sided die twice. What is the probability of rolling an even number and then an odd number? A) 1 16 B) 1 8 C) 1 4
jeyben [28]

<u>Answer:</u>

The probability of rolling an even number and then an odd number is \frac{1}{4} Option C is correct

<u>Solution:</u>

Given, You roll a six-sided die twice.  

We have to find what is the probability of rolling an even number and then an odd number?

We know that, rolling single die two times is equivalent to two dice rolling at a time.

\text { Now, probability of an event }=\frac{\text { number of favourable outcomes }}{\text { total possible outcomes }}

So, now, total possible outcomes = 6 x 6 = 36

And, number of favourable outcomes = 3 even on 1st die x 3 odd on 2nd die = 3 x 3 = 9

Then, probability = \frac{9}{36} = \frac{1}{4}

Hence, the probability of given condition is \frac{1}{4}  and option c is correct.

5 0
3 years ago
Read 2 more answers
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