All categories

3 years ago

4/10 + 1/2 please help-

Mathematics

2 answers:

Mazyrski [523]3 years ago

7 0

Exact Form:

9 /10

Decimal Form:

0.9

Hope this helps :)

SOVA2 [1]3 years ago

4 0

Answer:

4/10 + 1/2 would be 9/10 because 1/2 of 10 is 5 so 5/10 +4/10 would equal 9/10 hope that helps :)

Step-by-step explanation:

You might be interested in

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago

-1 1/5 + 3/4 I’m bad with fractions and don’t know how to do these it’s due tomorrow

Dovator [93]

-1 1/5 + 3/4 = -0.45

1/5 is 0.2 in decimal form, and 3/4 is 0.75:

-1.2 + 0.75 = -0.45

Hope this helps!

8 0

3 years ago

Read 2 more answers

26.45+4.79+120.02-3.20=

viva [34]

The answer to your question is = 148.06

6 0

3 years ago

Read 2 more answers

Eric randomly surveyed 150 adults from a certain city and asked which team in a contest they were rooting for, either North Hig

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

A 95% confidence level interval will have 0.52 (lower interval) & 0.68 (upper interval) which means that that if 90 individuals root for North HS then p value is 0.6 which will fall in the 95% confidence interval range.
For the option B the p value will also be same as in case A hence B is true as an alternative hypothesis.
We can calculate P value

Confidence Interval = p ± z $\sqrt{p(1-p)/n}$

4 0

3 years ago

Look at the two expressions. 3x−18 3⋅(x−6) Are the two expressions equivalent?

Ivanshal [37]

Yes because if you distribute the 3 and multiply it with 3 • x and 3• 6 you’ll get 3x-18

8 0

2 years ago

Read 2 more answers