All categories

3 years ago

Please help me with this, thanks.

Mathematics

1 answer:

balu736 [363]3 years ago

7 0

$\frac{3^{-2} }{8^{-2}} \frac{3^{3} }{8^{3} } \frac{1^{3} }{3^{3} } \frac{1^{4} }{3^{4} } = \frac{3^{ (-2+3)} 1^{ (3+4)}}{8^{ (-2+3) }3^{(3+4)}} = \frac{3^{1}1^{7}}{8^{1} 3^{7} } = \frac{3}{8 * 3^{7} } = \frac{1}{8*3^{6} } = \frac{1}{5,832}$

You might be interested in

How do I solve this (show work)

amm1812

Lol free points :----------------)

7 0

4 years ago

What is the area of a triangle whose base measures 3.2 centimeters and whose height is 4.6 centimeters.

Assoli18 [71]

Answer:

7.36 cm²

Step-by-step explanation:

What is the area of a triangle whose base measures 3.2 centimeters and whose height is 4.6 centimeters.

The formula for the area of a triangle is given as:

1/2 × Base × Height

Base = 3.2 cm

Height = 4.6 cm

Hence,

Area of the triangle = 1/2 × 3.2 × 4.6

= 7.36 cm²

4 0

3 years ago

The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with a

Lady_Fox [76]

Answer:

1) The most typical 68% of pregnancies last between 250 and 282 days, the most typical 95% between 234 and 298 days, and the most typical 99.7% between 218 and 314 days.

2) 15.87% of all pregnancies last less than 250 days

2a) 83.5% of pregnancies last between 241 and 286 days

2b) 10.56% of pregnancies last more than 286 days.

2c) 0% of pregnancies last more than 333 days

3) A pregnancy length of 234.6 days cuts off the shortest 2.5% of pregnancies.

4) The first quartile of pregnancy lengths is of 255.2, and the third quartile is of 276.8 days.

5) The most typical 72% of all pregnancies last between 248.72 and 283.28 days.

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 266 days and a standard deviation of 16 days.

This means that $\mu = 266, \sigma = 16$

(1) Using the 68-95-99.7% rule, between what two lengths do the most typical 68% of all pregnancies fall 95%, 99.7%?

68%: within 1 standard deviation of the mean, so 266 - 16 = 250 days to 266 + 16 = 282 days.

95%: within 2 standard deviations of the mean, so 266 - 32 = 234 days to 266 + 32 = 298 days.

99.7%: within 3 standard deviations of the mean, so 266 - 48 = 218 days to 266 + 48 = 314 days.

The most typical 68% of pregnancies last between 250 and 282 days, the most typical 95% between 234 and 298 days, and the most typical 99.7% between 218 and 314 days.

(2) What percent of all pregnancies last less than 250 days?

The proportion is the p-value of Z when X = 250. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{250 - 266}{16}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.1587*100% = 15.87%.

15.87% of all pregnancies last less than 250 days.

(a) What percentage of pregnancies last between 241 and 286 days?

The proportion is the p-value of Z when X = 286 subtracted by the p-value of Z when X = 241. So

X = 286

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{286 - 266}{16}$

$Z = 1.25$

$Z = 1.25$ has a p-value of 0.8944.

X = 241

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{241 - 266}{16}$

$Z = -1.56$

$Z = -1.56$ has a p-value of 0.0594.

0.8944 - 0.0594 = 0.835*100% = 83.5%

83.5% of pregnancies last between 241 and 286 days.

(b) What percentage of pregnancies last more than 286 days?

1 - 0.8944 = 0.1056*100% = 10.56%.

10.56% of pregnancies last more than 286 days.

(c) What percentage of pregnancies last more than 333 days?

The proportion is 1 subtracted by the p-value of Z = 333. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{333 - 266}{16}$

$Z = 4.19$

$Z = 4.19$ has a p-value of 1

1 - 1 = 0% of pregnancies last more than 333 days.

(3) What length cuts off the shortest 2.5% of pregnancies?

This is the 2.5th percentile, which is X when Z = -1.96.

$Z = \frac{X - \mu}{\sigma}$

$-1.96 = \frac{X - 266}{16}$

$X - 266 = -1.96*16$

$X = 234.6$

A pregnancy length of 234.6 days cuts off the shortest 2.5% of pregnancies.

(4) Find the quartiles for pregnancy length.

First quartile the 25th percentile, which is X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 266}{16}$

$X - 266 = -0.675*16$

$X = 255.2$

Third quartile is the 75th percentile, so X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 266}{16}$

$X - 266 = 0.675*16$

$X = 276.8$

The first quartile of pregnancy lengths is of 255.2, and the third quartile is of 276.8 days.

(5) Between what two lengths are the most typical 72% of all pregnancies?

Between the 50 - (72/2) = 14th percentile and the 50 + (72/2) = 86th percentile.

14th percentile:

X when Z = -1.08.

$Z = \frac{X - \mu}{\sigma}$

$-1.08 = \frac{X - 266}{16}$

$X - 266 = -1.08*16$

$X = 248.72$

86th percentile:

X when Z = 1.08.

$Z = \frac{X - \mu}{\sigma}$

$1.08 = \frac{X - 266}{16}$

$X - 266 = 1.08*16$

$X = 283.28$

The most typical 72% of all pregnancies last between 248.72 and 283.28 days.

7 0

3 years ago

Evaluate the expression 2⋅(3^6⋅3^−5) a) 2/3 b) 3 c) 6 d) 18

frosja888 [35]

D be it’s all the great number for it

6 0

3 years ago

Read 2 more answers

How do i solve this?

bixtya [17]

Answer:

$5.4 cm^{2}$

Step-by-step explanation:

The area of triangle is given by 0.5bh where b is the base while h is height. Therefore, A=0.5bh

For triangle ABC

15=0.5*5*h

H=6cm

Using the concept of similarity and enlargement

$\frac {B_a}{B_d}=\frac {H_a}{H_d}$

Where B and H represent base and height respectively. Subscripts a and d represent triangle ABC and DEF.

Therefore

$\frac {5}{3}=\frac {6}{H_d}\\H_d=3.6\ cm$

Area will be

$A=0.5*3*3.6=5.4\cm^{2}$

4 0

4 years ago