All categories

3 years ago

Rewrite the expression in the form 2". E

Mathematics

1 answer:

Vlad [161]3 years ago

6 0

Answer:

z^ {8/5}

Step-by-step explanation:

$(z^{\frac{-4}{3})^{\frac{-6}{5} } \\We\ know\ that\ a^{(b)^c} = a^{bc}\\\\Hence, \\\\{z^{\frac{-4}{3})^{\frac{-6}{5}}\\}={z^{\frac{-4}{3}*\frac{-6}{5}\\}\\=z^{\frac{24}{15}}\\=z^\frac{8}{5}$

You might be interested in

Crystal bought new art supplies that were on sale for $25.45. If

Keith_Richards [23]

Answer:

The orginal price was $33.09.

Step-by-step explanation:

You take 30% and move the decimal two places to the left, which gives you .3

Then mutiply .3 to $25.45. This gives you $7.64.

Add $7.64 to discounted price ($25.45) and you get $33.09.

5 0

3 years ago

Based only on the given information, it is guaranteed that AC = BC.

Alenkasestr [34]

False :) hope it help

5 0

3 years ago

Read 2 more answers

A farmer owns 70 acres of land. Of the 70 acres, only 65% can be farmed. How many acres are available for farming? acres are ava

denis-greek [22]

45.5 acres of land can be farmed and 24.5 cannot be farmed.

4 0

2 years ago

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo

muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: $P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}$

Where $\lambda$ is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

$P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}$

Substitute the respective values in the above formula.

$P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})$

$P(k > 2)= 1 - (0.44596)$

$P(k > 2)=0.55404$

Hence, the required probability is 0.55404.

4 0

3 years ago

How is multiplying fractions the same as multiplying whole numbers and different

gtnhenbr [62]

Well one similarity is that since you have to multiply, integer rules still apply and that you can also multiply variables. A difference is that with fractions, you have to make a number into an improper fraction rather than just the regular way which would most likely be the whole number with a decimal.

5 0

3 years ago

Read 2 more answers