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erma4kov [3.2K]
3 years ago
13

How many phone numbers are possible for one area code if the first four numbers are 202-1 , in that order , and the last three n

umbers are 1-7-8 in any order
Mathematics
1 answer:
Mkey [24]3 years ago
3 0

6 phone numbers are possible for one area code if the first four numbers are 202-1

<u>Solution:</u>

Given that, the first four numbers are 202-1, in that order, and the last three numbers are 1-7-8 in any order  

We have to find how many phone numbers are possible for one area code.

The number of way “n” objects can be arranged is given as n!

Then, we have three places which changes, so we can change these 3 places in 3! ways

n! = n \times (n - 1)!

Hence 3! is found as follows:

3! = 3 \times (3-1)!\\\\3! = 3 \times 2!\\\\3! = 3 \times 2 \times 1 = 6

So, we have 6 phone numbers possible for one area code.

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Answer:

I think the correct answer is option b.

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Need help with this please.
iogann1982 [59]
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4 0
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Omar rented a truck for one day. There was a base fee of $17.95, and there was an additional charge of 98 cents for each mile dr
stiv31 [10]

The given question is incomplete. The complete question is:

Omar rented a truck for one day. There was a base fee of $17.95, and there was an additional charge of 98 cents for each mile driven. Omar had to pay $23 when he returned the truck. For how many mile did he drive the truck?

Answer: Omar drove the truck for 5.15 miles

Step-by-step explanation:

Base fee = 17.95 $

Additional charge per mile = 98 cents = 0.98 $     ( 100cents = 1$)

Now Omar payed = 22 $

Let the miles he travelled = x

Now , 17.95+0.98\times x=23

Solvimg for x :

x=5.15miles

Thus Omar drove the truck for 5.15 miles

5 0
3 years ago
Neo is making invitations for their birthday party.
VMariaS [17]

Answer:

15%

Step-by-step explanation:

40=100%

20=50%

4=10%

2=5%

10%+5%=15%

4+2=6

6 0
3 years ago
The number of hours worked per year per person in a state is normally distributed with a standard deviation of 39. A sample of 1
evablogger [386]

Answer:

The 98% confidence interval for the population mean number of hours worked per year per person is (2146, 2193).

Step-by-step explanation:

The question is incomplete.

The number of hours registered in the sample are:

2051 2061 2162 2167 2169 2171

2180 2183 2186 2195 2196 2198

2205 2210 2211

The sample mean can be calculated as:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{15}(2051+2061+2162+2167+2169+2171+2180+2183+2186+2195+2196+2198+2205+2210+2211)\\\\\\M=\dfrac{32545}{15}\\\\\\M=2169.67\\\\\\

We have to calculate a 98% confidence interval for the mean.

The population standard deviation is know and is σ=39.

The sample mean is M=2169.67.

The sample size is N=15.

As σ is known, the standard error of the mean (σM) is calculated as:

\sigma_M=\dfrac{\sigma}{\sqrt{N}}=\dfrac{39}{\sqrt{15}}=\dfrac{39}{3.873}=10.07

The z-value for a 98% confidence interval is z=2.326.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_M=2.326 \cdot 10.07=23.43

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 2169.67-23.43=2146\\\\UL=M+t \cdot s_M = 2169.67+23.43=2193

The 98% confidence interval for the population mean is (2146, 2193).

3 0
3 years ago
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