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GREYUIT [131]
3 years ago
9

Help me with this math question ASAP please!!

Mathematics
2 answers:
Artyom0805 [142]3 years ago
8 0

hello! i'm not sure what the answer is but my recommendation is to download "photo math" you take a pic of your math problem and get answer answer! hope this helped !

stich3 [128]3 years ago
3 0
Heres the answer hope this helped you out

You might be interested in
A city planner wants to build a road parallel to 2nd Ave . What is the slope of the new road?
ludmilkaskok [199]

Answer:

0

Step-by-step explanation:

By inspection or finding points, we see that 2nd ave is perfectly flat, or in other words, has slope of 0.

If line 1 is parallel to line 2, it means their slopes are the same.

So, the new road parallel to 2nd ave has a slope of \boxed{0}.

6 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
A 24 ft ladder is positioned next to a house for window cleaning. The manufacturer recommends the ladder rest at an angle of 76°
Butoxors [25]

Answer:

6 feet

Step-by-step explanation:

cos = adj/hyp

cos76 = adj/24

multiply both sides by 24

24 * cos76 = adj

5.806125494392025 = adj

Rounded

6 feet

8 0
3 years ago
PLEASE HELP WITH THIS
kobusy [5.1K]
I think they are not always the best deal, but most are. Sometimes they are not good if they opponent has a even lower price then them.
8 0
3 years ago
Read 2 more answers
CAN SOMEONE HELP ME WITH THIS?!
lutik1710 [3]

Answer:

a = 14

b = 24

c = 24.9

A = 33.2 degrees

B = 70 degrees

C = 76.8 degrees

Step-by-step explanation:

a/sin(A) = b/sin(B) = c/sin(C)

14/sin(A) = 24/sin(70)

sin(A)×24 = sin(70)×14

sin(A) = sin(70)× 14/24 = sin(70) × 7/12 = 0.548154029...

A = asin(0.548154029...) = 33.240464... degrees

the sum of all angles in a triangle is airways 180 degrees.

C = 180 - 70 - 33.240464... = 76.75954... degrees

24/sin(70) = c/sin(76.75954...)

c = 24×sin(76.75954...)/sin(70) = 24.86133969...

6 0
3 years ago
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