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Gemiola [76]
3 years ago
11

Find the unit price if 7 pounds of sugar sells for $2.50

Mathematics
2 answers:
ohaa [14]3 years ago
7 0
2.50 divide by 7= .36 cents for each pound
SpyIntel [72]3 years ago
7 0

Answer: 36 cents

Step-by-step explanation: Unit price means the cost per unit which in this case is the cost per pound of sugar.

Since we know that it costs $2.50 for 7 pounds of sugar, to find the cost for 1 pound of sugar, we need to divide 7 into $2.50 or just 2.50 ÷ 7.

2.50 ÷ 7 = 0.357142857 which we can round up to 36 cents.

Therefore, the unit price is 36 cents per pound of sugar.

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Solve for e.<br><br> 4/3=-6e-5/3
cricket20 [7]

Answer:

e = 1/2

Step-by-step explanation:

to solve foe e in  this 4/3=-6e-5/3

solution

4/3=-6e-5/3

4/3 + 5/3 = 6e

find the lcm of the left hand side

4 + 5/3 = 6e

9/3 = 6e

cross multiply

3 x 6e = 9 x 1

18e = 9

divide both sides by the coefficient of e which is 18

18e /18 = 9/18

e = 1/2

therefore the value of e in the expression above is evaluated to be equals to 1/2

5 0
3 years ago
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
3 years ago
Which of the following functions are an example of exponential decay???
Yanka [14]

Answer:

C. II only

Step-by-step explanation:

iyzgkxhldlufulduo

7 0
3 years ago
It costs $12.50 to see a movie at a theater near Jason’s house. It costs only $1.95 to rent a DVD from an online service. A DVD
dolphi86 [110]
The answer would be $11.8 then just round that up to the nearest whole number which would be $12. Hope this helped if I’m incorrect I apologize:/but I’m pretty sure that’s right!

5 0
3 years ago
Please answer this question ‍♀️
krek1111 [17]

Answer:

sum is 1,2,4,8,16,32,64,128,256,512

5 0
2 years ago
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