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Sphinxa [80]
3 years ago
13

Is 7,849 a reasonable answer for 49 x 49?

Mathematics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

2,401

Step-by-step explanation:

No because 49x49 is 2,401

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Can someone help me please​
Ad libitum [116K]

Answer:

a) = 8

Step-by-step explanation:

You add up all the numbers then divide that number by how many numbers you had. 4+5+7+11+13=40 40/5.

Do you want the rest?

3 0
3 years ago
Read 2 more answers
Which is f(5) for the function -2x2 + 2x - 3?
Schach [20]

The value of f(5) is -37

<em><u>Solution:</u></em>

<em><u>Given function is:</u></em>

f(x) = -2x^2+2x+3

We have to find the value of f(5)

To find the value of f(5), we have to substitute x is equal to 5 in given function

Plug in x = 5 in f(x)

f(x) = -2x^2+2x+3\\\\f(5) = -2(5)^2+2(5) + 3\\\\\text{Simplify the above expression }\\\\f(5) = -2(25) + 10 + 3\\\\\text{Solve the above expression }\\\\f(5) = -50 + 10 + 3\\\\f(5) = -40 + 3 = -37

Thus value of f(5) is -37

6 0
3 years ago
a particular species of bamboo can grow up to 91 centimeters in a day. how fast is this in inches per hour
solniwko [45]
I would probably just do 91 times 60 since 60 min is 1 hour
8 0
3 years ago
If tom has 5 time as many as read and read have 5 how many is that
8_murik_8 [283]
So if he has 5 time more than he has read(5) than 5 multiplied by 5 is 25. I hope this helps 
6 0
4 years ago
Read 2 more answers
A simple random sample of size nequals57 is obtained from a population with muequals69 and sigmaequals2. Does the population nee
dusya [7]

Answer:

The population does not need to be normally distributed for the sampling distribution of \bar{X} to be approximately normally distributed. Because of the central limit theorem. The sampling distribution of \bar{X} is approximately normal.

Step-by-step explanation:

We have a random sample of size n = 57 from a population with \mu = 69 and \sigma = 2. Because n is large enough (i.e., n > 30) and \mu and \sigma are both finite, we can apply the central limit theorem that tell us that the sampling distribution of \bar{X} is approximatelly normally distributed, this independently of the distribution of the random sample. \bar{X} is asymptotically normally distributed is another way to state this.

6 0
3 years ago
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