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amm1812
3 years ago
12

How many unique ways are there to arrange the letters in the word PRIOR?

Mathematics
1 answer:
horrorfan [7]3 years ago
7 0
PRIOR MEAN PREVIOUS SO THAT IS UR ANSWER CUH a b c d the answer is yes
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Mary has $65,000 to invest in two accounts. The first account is a private savings offering 6% annual interest and the second ac
const2013 [10]

Answer:

first account with 6% interest ( x ) = $46000

second account with 4% interest ( y ) = $19000

Step-by-step explanation:

Principal = $65000

first interest rate = 6%

second interest rate = 4%

assuming ; X = money invested in first account ,  Y = money invested in second account

hence :

( 65000 - y ) * 6% + ( 65000 - x ) * 4% = 3520

3900 - 0.06 y  + 2600 - 0.04 x = 3520

0.04 x + 0.06 y = 3900 + 2600 - 3520

0.04 x + 0.06 y = 2980 ----- ( 1)

        x + y = 65000  ------- ( 2 )

solve equation 1 and 2 simultaneously by elimination method

0.04 x + 0.06 y = 2980 ------  ( 3 )

0.06 x + 0.06 y = 3900 ------- ( 4 )

subtract equation 3 form equation 4

0.02 x  = 920 ,   hence x = $46000

therefore y = $65000 - $46000 =  $19000

6 0
3 years ago
25 points | look at the image please
Anton [14]

Answer:

Ok I look now what

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

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

So, X ~ Normal(\mu=75, \sigma^{2} =20)

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 between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</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 between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</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 between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
2 years ago
Each leg of a 45°-45°-90° triangle measures 14 cm. What is the length of the hypotenuse? 7 cm cm 14 cm cm
Mashutka [201]
Check the picture below.

8 0
3 years ago
Read 2 more answers
If p(x) = 2x2 – 4x and q(x) = x – 3, what is (p*q) (x)
jonny [76]
Hello:
(poq)(x) = p(q(x))=  p(x-3) = 2(x-3)²-4(x-3) = 2(x²-6x+9)-4(x-3)
(poq)(x) = 2x²-12x+18-4x+12
(poq)(x) = 2x²-16x+30... (answer : C )

6 0
3 years ago
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