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3 years ago

True or False? An inscribed angle is formed by two radii.

Mathematics

1 answer:

strojnjashka [21]3 years ago

3 0

Answer:

False is the answer

An inscribed angle is not formed by two radii

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PLZ HELP!

AnnyKZ [126]

Answer:

$0.25

Step-by-step explanation:

Site A pays ...

... $0.65 × 5 = $3.25

Site B pays ...

... y = 0.60·5 = 3.00 . . . dollars

Site A pays more by ...

... $3.25 - 3.00 = $0.25

_____

You can also look at the difference in unit rates. Site A pays $0.65 per photo, while site B pays $0.60 per photo, a difference of $0.65 - 0.60 = $0.05 per photo. Then 5 photos will be worth $0.05 × 5 = $0.25 more at Site A.

4 0

3 years ago

What is the answer to 9x8 2/3

Ira Lisetskai [31]

The answer is 78. This is because you change 8 2/3 into a improper fraction which is 26/3. Then do 26/3×9/1. You should cross cancel so cancel out 3 and change it into 1 and change 9 into 3. So now your problem is 26×3=78.
~JZ
Hope it helps.

7 0

4 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 $t_{\alpha/2}=2.101$