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3 years ago

Please help ill give you brainliest

Mathematics

1 answer:

KIM [24]3 years ago

4 0

Answer:

Sean has 4 bags left over!

Step-by-step explanation:

Sean has 10 pounds of birdseed, which he separates into smaller bags. The line plot shows the weights of 8 bags Sean made. Sean's Bags of Birdseed x XXXXCO - 2 1 3 1 5 8 Weight (pounds) 3 4 1 co How many pounds of birdseed does Sean have left after making the bags? Move numbers to the boxes to show the answer. If there is no whole number, put a 0 in the first box.

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On any given day, the number of users, u, that access a certain website can be represented by the inequality 125-4530

Masteriza [31]

Answer:

Step-by-step explanation:

Our inequality is |125-u| ≤ 30. Let's separate this into two. Assuming that (125-u) is positive, we have 125-u ≤ 30, and if we assume that it's negative, we'd have -(125-u)≤30, or u-125≤30.

Therefore, we now have two inequalities to solve for:

125-u ≤ 30

u-125≤30

For the first one, we can subtract 125 and add u to both sides, resulting in

0 ≤ u-95, or 95≤u. Therefore, that is our first inequality.

The second one can be figured out by adding 125 to both sides, so u ≤ 155.

Remember that we took these two inequalities from an absolute value -- as a result, they BOTH must be true in order for the original inequality to be true. Therefore,

u ≥ 95

and

u ≤ 155

combine to be

95 ≤ u ≤ 155, or the 4th option

4 0

3 years ago

Antonio eats 1 slice of pizza in 3 minutes. How long will it take Antonio to eat 4 slices?

GarryVolchara [31]

Answer:

12 minutes

4*3

That's the answer

8 0

3 years ago

Is there anyone that can help me with this problem?

saul85 [17]

Answer:

your answer will be option 2

None of the choices are correct.

Step-by-step explanation:

have a nice day and good luck!!!

6 0

3 years ago

Read 2 more answers

Help with number 7!! You don’t have to do it but how do you do the problem??

musickatia [10]

80 * 5/8 will give you how many are thoroughbreds.

80 * 5/8 = 50

Now subtract this from all of the horses.

80 - 50 = 30.

There are 30 quarter horses.

OR If 5/8 of the horses are thoroughbreds, then 3/8 are quarter horses.

80 * 3/8 = 30.

6 0

3 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 = 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

3 years ago