All categories

2 years ago

What should be added to 3/4+-5/6+1/2 to get 0

Mathematics

1 answer:

monitta2 years ago

7 0

Answer:

-5/12

Step-by-step explanation:

3/4+-5/6+1/2 = 5/12

5/12 - 5/15 = 0

You might be interested in

the present age of the father is twice HD old is the age of its own three years hence if two years hence the age of the father w

NemiM [27]

Big sundi and more bigger with spice

6 0

2 years ago

An item is regularly priced at $45 . it is on sale for 60% off the regular price. how much (in dollars) is discounted from the r

wariber [46]

The discount is 60% of regular price, or we can write it as
discount = 60% × regular price

input the numbers
discount = 60% × regular price
discount = 60% × 45
discount = 60/100 × 45
discount = 2,700/100
discount = 27

The answer is $27

4 0

3 years ago

The average kinetic energy of the atoms in an object determines its _____.

Strike441 [17]

I am quite sure that it is B: Temperature

7 0

3 years ago

Read 2 more answers

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

3 years ago

Question 2 (Multiple Choice Worth 1 points)

Oliga [24]

A vertical line can only pass through the x axsis not the y unless it was at 0 then it would pass through all the y points. A diagonal line can pass through (-5,1) but not a vertical choose x=-5 if you want it to be a vertical line going through (-5,0)

3 0

2 years ago