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2 years ago

13

Janelle babysits and charges $5 per hour. Which equation can be used to find how many hours, h, she needs to babysit in order to

make $55.

1 answer:

anastassius [24]2 years ago

8 0

Answer:

11 is your corect awnser neess to be 20 characters

Step-by-step explanation:

5 times 11 is 55

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Two co- interior angles are in the ratio 5:7 find the angles

VARVARA [1.3K]

Answer:

75°

105°

Step-by-step explanation:

Two co-interor= 180°

5+7= 12

ratio12= 180°

<em>{</em><em>f</em><em>i</em><em>n</em><em>d</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>t</em><em>h</em><em>e</em><em>r</em><em> </em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>s</em><em>,</em><em> </em><em>5</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>7</em><em>)</em>

12=180°

5×180÷12 = 75°

To find 7, just subtract 180-75= 105°

3 0

3 years ago

Three lightship flash simultaneously at 6 00 a.m the first lightship flashes every 12 sec the second lightship every 30 sec and

Dmitrij [34]

They will all flash their lights together at 5:00 pm

6 0

2 years ago

If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calcul

Wittaler [7]

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

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.

3 failures every twenty weeks

This means that for 1 week, $\mu = \frac{3}{20} = 0.15$

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Then

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

$P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607$

$P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898$

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

6 0

2 years ago

Help Plz Will Upvote

kykrilka [37]

Not the best handwriting but I hope it helped!

4 0

2 years ago

Read 2 more answers

What is the quotient when the fraction 1/3 is divided by the whole number 6?

dsp73

Answer:

$\frac{1}{18}$

Step-by-step explanation:

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2 years ago