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2 years ago

Help me? please?????????

Mathematics

2 answers:

bagirrra123 [75]2 years ago

3 0

8/63 is the simplest form of the fraction

Andrews [41]2 years ago

3 0

Answer:

$\huge\boxed{\tt{\frac{8}{63} }}$

Step-by-step explanation:

$\displaystyle \frac{8}{9} * \frac{1}{7} \\\\= \frac{8*1}{9*7} \\\\= \frac{8}{63} \\\\\rule[225]{225}{2}$

Hope this helped!

You might be interested in

50 pts!!!! Luciana's laptop has 3,000 pictures. The size of the pictures is skewed to the right, with a mean of 3.7MB and a stan

skelet666 [1.2K]

Answer:

Part A: No, you cannot.

Part B: 0.4491

Explanation:

Part A:

The mean of samples of symmetrical (bell shaped) distributions follow a normal distribution pattern.

Thus, for symmetrical distributions you can use the z-score tables to make calculations that permit calculate probabilities for particular values.

For skewed distributions, in general, the samples do not follow a normal distribution pattern, except that the samples are large.

Since a sample of 20 pictures is not large enough, the answer to this question is negative: you cannot accurately calculate the probability that the mean picture size is more than 3.8MB for an SRS (skewed right sample) of 20 pictures.

Part B.

For large samples, the Central Limit Theorem will let you work with samples from skewed distributions.

Although the distribution of a population is skewed, the Central Limit Theorem states that large samples follow a normal distribution shape.

It is accepted that samples of 30 data is large enough to use the Central Limit Theorem.

Hence, you can use the z-core tables for standard normal distributions to calculate the probabilities for a random sample of 60 pictures instead of 20.

The z-score is calculated with the formula:

z-score = (value - mean/ (standard deviation).

$z=(x-\mu )/\sigma$

Here, mean = 3.7MB, value = 3.8MB, and standard deviation = 0.78MB.

Thus:

$z=(3.7M-3.8MB)/(0.78MB)=-0.128$

Then, you must use the z-score table to find the probability that the z-score is greater than - 0.128.

There are tables that show the cumulative probability in the right end and tables that show the cumulative probability in the left end of a normal standard distribution .

The probability that a z-score is greater than -0.128 is taken directly from a table with the cumulative probability in the left end. It is 0.4491.

4 0

3 years ago

Find S^20 if the series 1 + 1.1 ..... is a (a) arithmetic (b) geometric

quester [9]

I'm guessing S^20 is actually referring to the 20th partial sum $S_{20}$ , with

$S_{20}=1+(1+1\times0.1)+(1+2\times0.1)+\cdots+(1+18\times0.1)+(1+19\times0.1)$

in the arithmetic case, and

$S_{20}=1+1.1+1.1^2+\cdots+1.1^{18}+1.1^{19}$

in the geometric case.

(a) Combining like terms, we have

$S_{20}=20+0.1(1+2+\cdots+18+19)$

and invoking the formula

$1+2+\cdots+(n-1)+n=\dfrac{n(n+1)}2$

we end up with

$S_{20}=20+0.1\dfrac{19\times20}2=39$

(b) Multiply $S_{20}$ by 1.1, then subtract this from $S_{20}$ :

$1.1S_{20}=1.1+1.1^2+1.1^3+\cdots+1.1^{19}+1.1^{20}$

$S_{20}-1.1S_{20}=-0.1S_{20}=1-1.1^{20}$
$\implies S_{20}=\dfrac{1.1^{20}-1}{0.1}\approx57.275$

7 0

3 years ago

Choose some numbers to compare to 55 use = and <,> WHAT NUMBERS COMPARE TO 55

horrorfan [7]

There's literally unlimited answers but here are some simple ones.

5 x 11 = 55

55 = 55

55 > 54

1 < 55

3 x 2 < 55

3 0

3 years ago