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3 years ago

Hey can you help me out with this problem?

Mathematics

1 answer:

Reika [66]3 years ago

5 0

Answer:

Step-by-step explanation:

20/2=10 smores

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Please help me w/ this geometry question. <br><br> image attached.

allsm [11]

x = 180 - (77 + 32 + 49)

x = 180 - 158

x = 22°

6 0

2 years ago

A diver descends to -15 feet after 1 minute, -30 feet after 2 minutes, and -45 feet after 3 minutes. If the diver keeps descendi

Andreas93 [3]

The diver is obviously descending at a rate of 15ft/1min, as stated, all you have to do is multiply 12 by 15 and you should get your answer: 180

8 0

3 years ago

5. The average cost of airfare for a one-way ticket from Las Vegas to Los Angeles was $67

masha68 [24]

the cost of airfare on Friday is $141

<u>Step-by-step explanation:</u>

Here we have , The average cost of airfare for a one-way ticket from Las Vegas to Los Angeles was $67 for the days of Monday through Friday. The cost of airfare Monday through Thursday was $40, $44, $50, $60 respectively. We need to find What was the cost of airfare on Friday.Let's find out:

Let amount of Friday be x , So Average cost of airfare for a one-way ticket from Las Vegas to Los Angeles is given by :

⇒ $\frac{40+44+50+60+x}{5} =67$

⇒ $\frac{194+x}{5} =67$

⇒ $194+x =67(5)$

⇒ $x =335-194$

⇒ $x =141$

Therefore , the cost of airfare on Friday is $141

6 0

3 years ago

Come up with 4 numbers that will give you an average of 11.6.

Goshia [24]

Answer:

step 1) 3, 2, 1.6, and 5.

step 2) 3+2+1.6+5+5 =16.6

16.6/5 = 3.32

step 3) ????????? both have .6 ,,, IDK

Step-by-step explanation:

idk if this might help.

3+2+1.6+5 = 11.6

11.6 + 5 = 16.6

16.6/5 = 3.32

3 0

3 years ago

A new phone-answering system installed by a certain utility company is capable of handling ten calls every 10 minutes. Prior to

Mrrafil [7]

Answer:

0.0137

Step-by-step explanation:

Let X be the random variable that measures the number of incoming calls every ten minutes.

If the incoming calls to the system are Poisson distributed with a mean equal to 5 every 10 minutes, then the probability that there are k incoming calls in 10 minutes is

$\bf P(X=k)=\frac{e^{-5}5^k}{k!}$

If the phone-answering system is capable of handling ten calls every 10 minutes, we want to find

P(X>10), or the equivalent 1 - P(X≤ 10).

But

1 - P(X≤ 10)= 1 -(P(X=0)+P(X=1)+...+P(X=10)) =

$\bf 1-\left (\frac{e^{-5}5^0}{0!}+\frac{e^{-5}5^1}{1!}+...+\frac{e^{-5}5^{10}}{10!}\right)=\\=1-e^{-5}\left(\frac{5^0}{0!}+\frac{5^1}{1!}+...+\frac{5^{10}}{10!}\right)=1-0.9863=0.0137$

So, the probability that in a 10-minute period more calls will arrive than the system can handle is 0.0137

6 0

2 years ago