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3 years ago

In ΔXYZ, x = 5.2 inches, y = 9.8 inches and z=7.1 inches. Find the measure of ∠Z to the nearest degree.

Mathematics

2 answers:

lions [1.4K]3 years ago

5 0

Answer:45

Step-by-step explanation: Delta Math Said It Was The Answer

Hoochie [10]3 years ago

3 0

Answer:

Area≈23.365≈23.4

Step-by-step explanation:

Delta math

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What is the difference of the polynomials?<br><br> (8r6s3 – 9r5s4 + 3r4s5) – (2r4s5 – 5r3s6 – 4r5s4)

tangare [24]

8r^6s^3 - 9r^5s^4 + 3r^4s^5 - (2r^4s^5 - 5r^3s^6 - 4r^5s^4) =
8r^6s^3 - 9r^5s^4 + 3r^4s^5 - 2r^4s^5 + 5r^3s^6 + 4r^5s^4 =
8r^6s^3 -5r^5s^4 + r^4s^5 + 5r^3s^6 <==

6 0

3 years ago

Read 2 more answers

An employer has a staff of eighty actuaries, ten of whom are student actuaries. A student actuary is allowed a total of ten week

sergiy2304 [10]

Answer:

0.1111

Step-by-step explanation:

From the given information;

Number of staffs in the actuary = 80

Out of the 80, 10 are students.

i.e.

P(student actuary) = 10/80 = 0.125

number of weeks in a year = 52

off time per year = 10/52 = 0.1923

P(at work || student actuary) = (50 -10/52)

= 42/52

= 0.8077

P(non student actuary) = (80 -10)/80

= 70 / 80

= 0.875

For a non-student, they are only eligible to 4 weeks off in a year

i.e.

P(at work | non student) = (52-4)/52

= 48/52

= 0.9231

∴

P(at work) = P(student actuary) × P(at work || student actuary) + P(non student actuary) × P(at work || non studnet actuary)

P(at work) = (0.125 × 0.8077) + ( 0.875 × 0.9231)

P(at work) = 0.1009625 + 0.8077125

P(at work) = 0.90868

Finally, the P(he is a student) = (P(student actuary) × P(at work || student actuary) ) ÷ P(at work)

P(he is a student) = (0.125 × 0.8077) ÷ 0.90868

P(he is a student) = 0.1009625 ÷ 0.90868

P(he is a student) = 0.1111

5 0

3 years ago

The amount of money A accrued at the end of

Rudiy27

B is the answer no capppppp

7 0

2 years ago

An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68\%68%68, percent confidence

Komok [63]

Answer:

B. 10000 years old, with a margin of error of 400 years

Step-by-step explanation:

Assuming this complete problem: "12. An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a following could the spectrometer estimate as the age of the branch at the 95% confidence level?"

The possible options are:

A. 9500 years old, with a margin of error of 500 years

B. 10000 years old, with a margin of error of 400 years

C. 9500 years olds, with a margin of error of 50 years

D. 10000 years old, with a margin of error of 40 year

Solution to the problem

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error for this case is given by this formula:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

For the first case the confidence was 68% so then $\alpha =1-0.68 =0.32$ and $\alpha/2 =0.16$ we can find a critical value on the normal standard distribution that accumulates 0.16 of the area on each tail and this value is $z=\pm 0.994$ , because P(z<-0.944) = 0.16 and P(Z>.944) = 0.16

The margin of error can be expressed like this:

$ME = z_{\alpha/2} SE$

We can solve for the standard error on this case and we got:

$SE = \frac{ME}{z_{\alpha/2}} =\frac{200}{0.994}=201.207$

And then for the new confidence interval we need to calculate the new $z_{\alpha}$ . For the first case the confidence was 95% so then $\alpha =1-0.95 =0.05$ and $\alpha/2 =0.025$ we can find a critical value on the normal standard distribution that accumulates 0.025 of the area on each tail and this value is $z=\pm 1.96$ , because P(z<-1.96) = 0.025 and P(Z>1.96) = 0.025. So then the new margin of error would be: