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olasank [31]
2 years ago
7

Show that 5^133 +26 is a multiple of 31

Mathematics
2 answers:
MArishka [77]2 years ago
4 0

Answer:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45

Step-by-step explanation:

maks197457 [2]2 years ago
3 0

Answer:

100 PLS BRINLEY REPORT ME IF THE ANSWER WRONG

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Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
Pls help!!!!!!!!!!!!!​
Wewaii [24]

Answer:

All triangles have 180 degrees.

Step-by-step explanation:

7 0
3 years ago
What is the sign of -567.45 + 567.45−567.45
Cloud [144]

Answer:

The answer is (- 567.45)

8 0
3 years ago
In a sequnece u5=-3.7 and u15 =-52.3, what is the 19th term
PilotLPTM [1.2K]

u19=-71.74

Step-by-step explanation:

u5=a+4d=-3.7....(1)

u15=a+14d=-52.3....(2)

-10d=48.6

d=-48.6/-10

d=-4.86

Substitute-4.86 into....(1)

u5=a+4(-4.86)=-3.7

a+(-19.44)=-4.7

a=-3.7-(-19.44)

a=15.74

19term=a+18d=?

=15.74+18(-4.86)

=15.74+(-87.48)

19th term =-71.74

6 0
2 years ago
Lastima Nelson, Inc. is recruiting a new Chief Financial Officer (CFO). The recruiting budget is $44,000. Lastima has spent $10,
Musya8 [376]
The budget is $44000

Total spent so far is $10000 + $8500 = $18500

The amount left to spend = 44000 - 18500 = $25500

$25500 is the maximum 6% commission that Lastima Nelson Inc. can spend and stays within the budget

Let the maximum price of the home be x
This value of x will be the 100% before the 6% commission is calculated of it.

6% of x is 25500
1% of x is 25500 ÷ 6 = $4250
100% of x is 4250 × 100 = $425,000

So, the maximum value of the home is $425,000
5 0
3 years ago
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