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

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>

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:

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