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3 years ago

Please helpppppppppp

Mathematics

2 answers:

Lostsunrise [7]3 years ago

8 0

Answer:

see picture above shown

BigorU [14]3 years ago

6 0

Answer:

The answer is -a^2 + 3a +1

Let me know if it's right, this is what I got when I did the problem :)

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Find the value of x if you solve one please say the question number and you dony have to show work

zavuch27 [327]

Answer:

question 1-- x=15

question 2-- x=21

7 0

3 years ago

Read 2 more answers

Convert the given measurements to the new units to be 37cm

brilliants [131]

Answer:

37cm = 370 millimeters

37cm = 0.37 meter

37cm = 0.00037 kilometers

Step-by-step explanation:

$1cm = 10 millimeters \\\\1 cm = \frac{1}{100} meter\\\\1cm = \frac{1}{100000} kilometer\\\\$

37cm = 370 millimeters

37cm = 0.37 meter

37cm = 0.00037 kilometers

6 0

3 years ago

Select three numbers that are between -0.55 and -0.54.

pochemuha

-0.548, -0.5469, and -0.542

6 0

3 years ago

Find the missing part.

sveta [45]

Answer:

The answer is $a = \frac{15}{4}$

Step-by-step explanation:

We have to:

$cos(\theta) = \frac{adjacent\ side}{hypotenuse}$

So, if we take the 60 degree angle of the smallest triangle, we have to:

$cos(60) = \frac{a}{x}$

We do not know x. But if we take the sine of the angle of 30 degrees and the main triangle we have:

$sin(30) = \frac{x}{15}\\\\x =\frac{15}{2}$

Then:

$cos(60) = \frac{a}{\frac{15}{2}}$

$a = cos(60)\frac{15}{2}$

$a = \frac{15}{4}$

7 0

3 years ago

A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p

andreyandreev [35.5K]

Answer:

B) 0.894

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that $\mu = 0.55$

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577$

$P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894$

5 0

4 years ago