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Andrew [12]
3 years ago
15

Now, think about multiplying-25×-35.

Mathematics
1 answer:
Aliun [14]3 years ago
5 0

Answer:

Positive

Step-by-step explanation:

If you multiply both negative, it will always result to positive.

Rule:

+ × + = +

+ × - = -

- × + = -

- × - = +

You might be interested in
The equation 6x - 5y = 14 is written in standard form. Which point lies on
dimaraw [331]
Answer= H (-1, -4)

Explanation= 6(-1) - 5(-4) = 14
(-6) + (20) = 14
14=14 * it works !
GOODLUCK!
4 0
3 years ago
Pls and pls what is the easy way to solve maths problems or how can I solve maths problems with ease?
solong [7]

Answer:

Ok here is my pro tip:

Step-by-step explanation:

Use both halfs of you brain at the same time, causing you to passout and have your head hit the keyboard, giving you the correct answer

4 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
In zoe's class, 4/5 of the students have pets. of the students who have pets, 1/8 have rodents. what fraction of the students in
olasank [31]
Let us assume the number of student's in Zoe's class = x
Then
Number of students that have pets = 4x/5
Number of students that have rodents as pets = (1/8) * (4x/5)
                                                                          = x/10
Number of students in Zoe's class
that have pets that are not rodents = (4x/5) - (x/10)
                                                         = (8x - x)/10
                                                         = 7x/10
I hope that the procedure is clear enough for you to understand and this is the answer that you were looking for.
8 0
3 years ago
Read 2 more answers
What is the greatest common factor of 22 and 2?
Snowcat [4.5K]

Answer:

2

Step-by-step explanation:

2 is 2.

22 is 2•11.

The GCF is 2.

6 0
3 years ago
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