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yarga [219]
2 years ago
5

%7D%7Bhello%7D%7B%20%7B%7B%5Cboxed%7B%5Ccolor%7Baqua%7D%7B%20%20%5Cbinom%7B%3F%7D%7B%3F%7D%20%20%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D" id="TexFormula1" title="\small{\colorbox{black}{\red{\boxed{{ \sf\: \color{hotpink}{hello}{ {{\boxed{\color{aqua}{ \binom{?}{?} }}}}}}}}}}" alt="\small{\colorbox{black}{\red{\boxed{{ \sf\: \color{hotpink}{hello}{ {{\boxed{\color{aqua}{ \binom{?}{?} }}}}}}}}}}" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Ganezh [65]2 years ago
5 0
Black, hot pink, hello, ????
You might be interested in
3. Solve for y: -5x + 2y = 14
Schach [20]

Answer:

y= -1

Explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

8 0
3 years ago
Read 2 more answers
A can of soda weighs 336 grams. Find the weight in kilograms of 15 cans.
zaharov [31]

Answer:

5.04

Step-by-step explanation:

1 can equals to 336 grams

so 15 cans will equal to 336 X 15

which is 5040grams

now covert it into kg by dividing it by 1000

which 5.04

4 0
3 years ago
The búfalo children put all of their candy together and weighed it. It weighed 2 1/4 pound . They shared it amongst the five peo
ratelena [41]

Answer:

0.45 lbs

Step-by-step explanation:

2 1/4 can be converted to the decimal: 2.25

2.25÷5 is how you will set this problem up, because 2.25 lbs of candy is being divided amongst 5 people.

0.45 lbs is the answer

8 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
Solve using elimination.<br><br> 8x − 7y = –5<br> –x + 4y = –15
xz_007 [3.2K]
Steps
8x−7y=−5
Add 7y to both sides
8x−7y+7y=−5+7y
Simplify
8x=−5+7y
Divide both sides by 8
8x
8 =−
5
8 +
7y
8
Simplify
x=
−5+7y
8

7 0
3 years ago
Read 2 more answers
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