Multiply (-18)(-13) its for usatestprep

If the digits of a three-digit number are reversed in order, then the sum of the new resulting number and the original number co

Makovka662 [10]

First we have to setup system of equations for the given case.
Let x is the original number and y is the reverse three digit number.
Then,
x+y=665 Equation 1
x-y=297 Equation 2
Add both equations to get.
2x=962
Divide both sides by 2.
x=481
This is the original number. And reverse number is : 481+y=962
y=184

Here Hundredth digit is 4, and tenth digit is 8 which is 2 times of the hundredth digit.

Answer: Number is 481

6 0

2 years ago

2/ 3 (6 − 9x) = 16. Can someone pls help me

lidiya [134]

Answer:

-2 is your answer

Step-by-step explanation:

First, you get 2/3 and distribute it into the parentheses that gives you

4-6x=16

now you subtract 4 on both sides leaving you with

-6/-6 which cancels out you have

12/-6 whigh is -2 and that is your answer

8 0

3 years ago

Do the following set of points represent a function?Explain your answer (2,3),(3,8),(-1,4),(1,8)

mestny [16]

Answer:

Yes.

Step-by-step explanation:

Yes.

To have a function, you cannot use the same value for an x-coordinate more than once. Here, each value of an x-coordinate is used only once, so it is a function.

6 0

2 years ago

Read 2 more answers

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis

nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = the interval of time between the eruptions

So, X ~ N( $\mu=72, \sigma^{2} =23^{2}$ )

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 = 72 minutes

$\sigma$ = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 minutes is given by = P(X > 82 min)

P(X > 82 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{82-72}{23}$ ) = P(Z > 0.43) = 1 - P(Z $\leq$ 0.43)

= 1 - 0.6664 = 0.3336

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let $\bar X$ = sample mean time between the eruptions

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 = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{13} } }$ ) = P(Z > 1.57) = 1 - P(Z $\leq$ 1.57)

= 1 - 0.9418 = 0.0582

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let $\bar X$ = sample mean time between the eruptions

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 = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{34} } }$ ) = P(Z > 2.54) = 1 - P(Z $\leq$ 2.54)

= 1 - 0.9945 = 0.0055

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 minutes, then we conclude that the population mean must be more than 72, since the probability is so low.

6 0

3 years ago

Solve the following system of linear equations: 3x1+6x2+6x3 = -9 -2x1–3x2-3x3 = 3 If the system has infinitely many solutions, y

choli [55]

Answer:

The set of solutions is $\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}$

Step-by-step explanation:

The augmented matrix of the system is $\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]$ .

We will use rows operations for find the echelon form of the matrix.

In row 2 we subtract $\frac{2}{3}$ from row 1. (R2- 2/3R1) and we obtain the matrix $\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]$
We multiply the row 1 by $\frac{1}{3}$ .

Now we solve for the unknown variables:

$x_2+x_3=-7$ , $x_2=-7-x_3$
$x_1+2x_2+2x_3=-2$ , $x_1+2(-7-x_3)+2x_3=-2$ then $x_1=12$

The system has a free variable, the the system has infinite solutions and the set of solutions is $\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}$

6 0

2 years ago