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4 years ago

PLEASE HELP ME THANKS!

Mathematics

1 answer:

Marianna [84]4 years ago

4 0

Answer:

Step-by-step explanation:

C=2piR so 2(3.2)*3.14

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Angle 1 and Angle 2 are supplementary. Angle 1 is 20 degree less than nine times the size of angle 2.

nadya68 [22]

Answer:

∠1=160° and ∠2=20°

Step-by-step explanation:

Let ∠1 = x

∠2 = y

as these two angles are supplementary their sum is 180

that x+y=180 ----(A)

Also given that ∠1 is 20 degree less than nine times the size of ∠2.

Hence

x=9y-20

putting value of x in A and solving for y

9y-20+y=180

10y=180+20

10y=200

y=10

Putting this y in A

x+10=180

x=160

5 0

3 years ago

State the excluded values for each? 70v^2/100v

sleet_krkn [62]

Ok so any number tat makes the denomenator 0 or makes the inside of a square root negative is restricted
we only have a denomenaor so
100v=0
v=0
therefor 0 is the excluded value since 0/0 doesn't make sense

6 0

4 years ago

Three hikers are dividing the costs of a hiking trip evenly. They bought two permits that cost $15 each, three tram tickets that

cestrela7 [59]

Each hiker owes $41 or 4100 cents for the hiking trip.

Explanation

Cost of each permit is $15. So, the cost of two permits will be: $(\$15*2)= \$30$

Cost of each tram ticket is $19. So, the cost of three tickets will be: $(\$19*3)=\$ 57$

Cost of one can of bear spray is $36.

So, the total cost of the hiking trip $=\$30+\$57+\$36=\$123$

As the three hikers are dividing the total costs of the trip evenly, so each hiker owes $=\frac{\$123}{3}=\$41$ or 4100 cents.

3 0

3 years ago

A recent study found that the average length of caterpillars was 2.8 centimeters with a

pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

$\mu = 2.8, \sigma = 0.7$ .

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is one subtracted by the p-value of Z when X = 4, hence:

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

$Z = \frac{4 - 2.8}{0.7}$

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0

2 years ago

Can someone PLEASE help me on this Geometry question!!!

grandymaker [24]

What is your geometry question?

8 0

3 years ago