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3 years ago

I need help as soon as possible at 20 points

Mathematics

1 answer:

valina [46]3 years ago

4 0

9514 1404 393

Answer:

(15/14)² = 225/196

Step-by-step explanation:

Evaluating the expressions inside parentheses, we can see they are the same, eliminating a bit of work.

$(\frac{5}{7}\times\frac{3}{2})^5\div(\frac{5}{2}\times\frac{3}{7})^3=(\frac{15}{14})^5\div(\frac{15}{14})^3=(\frac{15}{14})^{5-3}=\boxed{(\frac{15}{14})^2=\frac{225}{196}}$

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mel-nik [20]

Answer:

x = 11.0

Step-by-step explanation:

b = √c2 - a2

= √132 - 6.88895043503172

= √121.54236190368

= 11.02463

5 0

4 years ago

Distribute -(-4v+6.3w-9.8)

Firdavs [7]

Answer:

4v-6.3w+9.8

Step-by-step explanation:

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Basically it's switching the signs

Hope that helped :)

3 0

3 years ago

Help pleaseeeeeeeeeee

Vladimir [108]

6x^2-6x+5 would be the correct answer

3 0

3 years ago

Which of the following equations is equivalent to x - y = 5?

Butoxors [25]

Y= x-5 is equivelent to the equation

4 0

3 years ago

Read 2 more answers

The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili

Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

$\lambda= 4.5$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

a) No accidents occur in one week.

$P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111$

b) 5 or more accidents occur in a week.

$P( x \geq 5) = 1-\displaystyle \sum P(x$

c) One accident occurs today.

The mean number of accidents per day is given by

$\lambda = \dfrac{4.5}{7} = 0.64$

$P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747$

5 0

3 years ago

I need help as soon as possible at 20 points ​

I need help as soon as possible at 20 points