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Evgen [1.6K]
3 years ago
12

What is the sum of 1 7/10 and 3 3/5?

Mathematics
2 answers:
Ksju [112]3 years ago
7 0

Answer:

5 3/10

Step-by-step explanation:

galina1969 [7]3 years ago
7 0

Answer:

5 3/10

-

You make the 3/5 into 6/10 and proceed with combining like terms

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On,Saturday, Alexa babysat her neighbors kids for 180 minutes. Then, on Sunday she babysat her little brother for 2 hours.how ma
patriot [66]

Answer:

5 hours

Step-by-step explanation:

To change minutes to hours you will have to divide the total minutes by 60 (because 60 minutes is in one hour), in this situation it would look like this:

180/60 = 3

So, Alexa babysat for 3 hours on Saturday.

Now, just add the 3 hours from Saturday and 2 hours from Sunday and you get a total of 5 hours!

<em>I hope this helps!!</em>

<em>- Kay :)</em>

8 0
3 years ago
38.7÷1.29 help I'm stuck
Darya [45]

1.29 can go into 38.7 30 times, Your answer is 30.

5 0
3 years ago
Read 2 more answers
Can anyone solve this asap? i’m completing a final.
Allisa [31]

Answer:

Step-by-step explanation:

The right option es a)

because the general form is:

ax^2+bx+c

so you can easyly identify b=-5

7 0
3 years ago
3x+ 32 +7x - 22 = 180
Sedaia [141]
Add 3x and 7x because they’re on the same and when you added it you’ll get 10x and than do 32+-22 and you’ll get 10 because 32-22 is 10 and you subtract the 10 to 180 and you’ll get 10x=170 and divide 170 by 10x to get x by itself when you divided you’ll get x=17 and that’s the answer.
6 0
3 years ago
If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distributio
goldenfox [79]

Answer:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And we want a normal distribution for the sample mean

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

X_i \sim N(\mu , \sigma), i=1,2,...,n

Because if we find the parameters we got:

E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu

Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}

And the deviation would be:

Sd (\bar X) = \frac{\sigma}{\sqrt{n}}

And we satisfy the condition:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

3 0
3 years ago
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