What to do when forget the pasword to the account?

Phillip is rolling two number cubes. how many different ways can he roll the number cubes and get a sum of 5?

Greeley [361]

4+1, 3+2, 2+3, 1+4. 4 total combinations, unless 4+1 and 1+4 are the same. In that case, there are only 2, 1+4, and 2+3

8 0

2 years ago

M in

klio [65]

Answer:

The copy machine makes 87.5 copies in one minute.

Step-by-step explanation:

Set up an equation to divide 4 into 350. This is because in order to know how many copies is being made in 1 minute, you need to determine the rate of copies per minute.

350 divided by 4 is 87.5.

3 0

2 years ago

Will give BRAINLEST :)

DiKsa [7]

Answer:

A

Step-by-step explanation:

This explanation mostly depends on what you're learning right now. The first way would be to convert this matrix to a system of equations like this.

g + t + k = 90

g + 2t - k = 55

-g - t + 3k = 30

Then you solve using normal methods of substitution or elimination. It seems to me that elimination is the quickest method.

g + t + k = 90

-g - t + 3k = 30

____________

0 + 0 + 4k = 120

4k = 120

k = 30

No you can plug this into the first two equations

g + t + (30) = 90

g + t = 60

and

g + 2t - (30) = 55

g + 2t = 85

now use elimination again by multiplying the first equation by -1

g + 2t = 85

-g - t = -60

_________

0 + t = 25

t = 25

Now plug those both back into one of the equations. I'll just do the first one.

g + (25) + (30) = 90

g = 35

Therefore, we know that Ted spent the least amount of time on the computer.

The second method is using matrix reduction and getting the matrix in the row echelon form, therefore solving using the gauss jordan method. If you would like me to go through this instead, please leave a comment.

3 0

2 years ago

Freyda has a circular rug with a radius of 120 centimeters. The fringe on the outside edge of the rug needs to be replaced. What

Gnom [1K]

Answer:

The Fringe of the rug is 754 cm.

Step-by-step explanation:

Given:

radius = 120 cm

We need to find the fringe of the outside rug.

Solution:

Since the rug is in the circular form.

We can say that fringe of the outside edge of the rug can be equal to circumference of the circle.

Then we will find the Circumference of the circle.

Circumference of the circle is given 2 times 'π' times radius 'r'.

framing in equation form we get;

Circumference of the circle = $2\pi r$

Circumference of the circle = $2 \times \pi \times 120 = 753.98 \approx 754\ cm$

Hence the Fringe of the rug is 754 cm.

8 0

2 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 = the interval of time between the eruption

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 = 0.3264

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 $\bar X$ = sample time intervals between the eruption

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 = 0.0526

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 $\bar X$ = sample time intervals between the eruption

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 = 0.0222

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