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sergij07 [2.7K]
2 years ago
15

PLEASEEE HELP I JUST NEED TO KNOW WHICH ONES ARE TRUE AND FALSE

Mathematics
2 answers:
liraira [26]2 years ago
6 0
1.) FALSE
2.) FALSE
3.)TRUE
4.)FALSE
Can someone pls clarify my answer for 4? Im kinda unsure ://
NARA [144]2 years ago
3 0

Answer:

Step-by-step explanation:

1-true

2-false

3-true

4-false

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mary had 1 3/4 pound of blackberries. Mary's friend ate 3/4 of the blackberries. How many pounds of blackberries did Mary friend
OLEGan [10]

Hey there! I'm happy to help!

If Mary's friend at 3/4 of blackberries, we will simply multiply the total number of blackberries by 3/4 to find how many blackberries Mary ate.

3/4×1 3/4= 1 5/16

Therefore, Mary's friend at 1 5/16 pounds of blackberries.

I hope that this helps! Have a wonderful day!

3 0
3 years ago
Answer me please now
Anna35 [415]

Answer:

1. $8.

2. 48 counters.

Step-by-step explanation:

1. A pair of socks cost 20 - 2(6)

= $8.

2. let the number of counters in Box A and Box be be x.

Then the number of counters in Box B after the transfer is x + 1/2 x.

x + 1/2 x = 72

1.5 x = 72

x = 72 / 1.5

= 48.

4 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 = <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
Nvm I don’t need anymore thanks anyways
solniwko [45]

Answer:

•________•♡♡♡•______•

6 0
3 years ago
I️ have travelled around 75% of the edge of a circular path for 10 miles. What is the diameter of the circular path
KonstantinChe [14]
The circumference of a circle is equal to pi * diameter

So 10 miles = 0.75 * pi * d

10/(0.75*pi) = d

d = <span>4.244 miles</span>
5 0
2 years ago
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