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faust18 [17]
3 years ago
10

Four minus 2X equals 16

Mathematics
2 answers:
olchik [2.2K]3 years ago
6 0

Answer:

-6

Step-by-step explanation:

4-2x=16

-4       -4

--------------

 -2x=12

/////////////

-2      -2

-----------

x=-6

marta [7]3 years ago
4 0
X=-6
4-2x=16
-4. -4
-2x=16
—— —-
-2. -2
x=-6
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Decrease 1440 by 80%
mario62 [17]

Answer:

288 will be ur ans

Step-by-step explanation:

1440 x 80/100

=1152

1440-1152

=288

5 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
What is the equation of the line graphed below? Answer choices shown in image as well as the graph
lorasvet [3.4K]

Answer:

A

Step-by-step explanation:

4 0
2 years ago
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In investing $6,200 of a couple's money, a financial planner put some of it into a savings account paying 4% annual simple inter
ddd [48]

Answer:

$ 2,600 was invested at 4% and $ 3,600 was invested at 9%.

Step-by-step explanation:

Given that in investing $ 6,200 of a couple's money, a financial planner put some of it into a savings account paying 4% annual simple interest, and the rest was invested in a riskier mini-mall development plan paying 9% annual simple interest, and the combined interest earned for the first year was $ 428, to determine how much money was invested at each rate, the following calculation must be performed:

3000 x 0.04 + 3200 x 0.09 = 408

2500 x 0.04 + 3700 x 0.09 = 433

2600 x 0.04 + 3600 x 0.09 = 428

Therefore, $ 2,600 was invested at 4% and $ 3,600 was invested at 9%.

7 0
3 years ago
−4x − 5y − z = 18<br> −2x − 5y − 2z = 12 <br> −2x + 5y + 2z = 4
MrRissso [65]

Answer: (x,y,z) = (-4,0,-2)

Step-by-step explanation:

6 0
3 years ago
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