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3 years ago

Im so lost what the answer is

Mathematics

1 answer:

Rainbow [258]3 years ago

5 0

Answer:

Unfortunately, we can't do anything with 41 as it has no factors. Since $3\sqrt{41}$ is already simplified, try submitting that as your answer. If that doesn't work then try un-doing their steps and get rid of the 3:

$3\sqrt{41}$

$\sqrt{41(9)}$

$\sqrt{369}$

Hope this helps!

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Log (w2+21) - log (10)

UNO [17]

Answer:

$w =7\ or\ w =3$

Step-by-step explanation:

Given

The right question is:

$\log(w^2+ 21) = \log(10w)$

Required

Solve for w

$\log(w^2+ 21) = \log(10w)$

Cancel log on both sides

$w^2+ 21 = 10w$

Equate to 0

$w^2 - 10w+ 21 = 0$

Split

$w^2 - 3w - 7w+ 21 = 0$

Factorize

$w(w - 3) - 7(w-3) = 0$

Factor out w - 3

$(w - 7)(w-3) = 0$

Solve for w

$w - 7 = 0\ and\ w - 3 = 0$

$w =7\ or\ w =3$

6 0

3 years ago

What is 2. 84 repeating as a mixed number in simplest form

Shalnov [3]

The answer would be 2 21/25.

8 0

4 years ago

PLEASE HELP ASAP!!!! YOU WILL GET 50 POINTS:)

slava [35]

Answer:

The remainder is 5

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

What value for z makes this equation true?

Paladinen [302]

(10×10)+(10×5)=10z
100+50=10z
150=10z
z=15 which is C. Hope it help!

4 0

3 years ago

Read 2 more answers

Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site. Please show your wor

natita [175]

Answer:

a) $\bar X=\frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

b) $s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

c) For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

d) $1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

Step-by-step explanation:

For this case we ave the following data:

1245 1245 1321 1191 1295 1330 1239 1250 1228

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{1245+1245+1321+1191+1295+1330+1239+1250+1228}{9}= 1260.444$

Part b

For this case we can use the following formula in order to find the sample standard deviation:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s = 45.580$

Part c

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".

For this case since the sampel size is n=9 <30 we can use the t distribution in order to find the confidence interval.

Part d

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

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

We need to find the degrees of freedom given by:

$df = n-1 = 9-1=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=1-0.95=0.05$ 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(0.025,8)".And we see that $t_{\alpha/2}=2.306$

Now we have everything in order to replace into formula (1):

$1260.444-2.306 \frac{45.580}{\sqrt{9}}= 1225.408$

$1260.444+2.306 \frac{45.580}{\sqrt{9}}= 1295.480$

8 0

4 years ago