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3 years ago

Whats the distance between (8,-3) and (4,-7)

Mathematics

1 answer:

Shtirlitz [24]3 years ago

7 0

Answer:

d=5.656854

Step-by-step explanation:

d=√ ( 4 − 8 ) 2 + (− 7 − (− 3 ) ) 2

d=√ ( − 4 ) 2 + ( − 4 ) 2

d= √ 16 + 16

d= √32

d=5.656854

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If a normal distribution has a mean of 44 and a standard deviation of 8, what is the zscore for a value of 50

Misha Larkins [42]

Answer:

0.75

Step-by-step explanation:

Z-score tells the no. of standard deviations a data point is from the mean.

the formula for z score is given by

z-score= (data value - mean) / standard deviation

Given:

mean=44

standard deviation= 8

data point=50

putting the values of mean(44), standard deviation(8) and given data value(50) in the above equation

z-score= (50-44)/8

= 6/8

=0.75 !

5 0

2 years ago

Bab in need of help!

katrin [286]

Answer:

Third option is the correct answer

Step-by-step explanation:

Given end points are: $(x_1, \:y_1) \:\&\: (x_2, \:y_2)$

$\huge\purple {\therefore Midpoint =\bigg(\frac{x_1+x_2}{2},\: \frac{y_1+y_2}{2}\bigg)}$

3 0

2 years ago

20 PTS<br> If a new car payment is $616.67 each month, what is the total car payment for one year?

vitfil [10]

Answer:

7400.04

Step-by-step explanation:

Well, if one month is 616.67, then one year is just 12 months. Multiply 6l6.67 by 12, and the answer is 7400.04

8 0

2 years ago

Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

7 0

3 years ago

5x + 3y = -3 <br><br> 2X - 3y = 24

gregori [183]

Answer:

Y=-5/3x-1

y=2/3x-8

If you are adding then it is:

7x=21

x=3

Step-by-step explanation:

5x+3y=-3

3y=-5x-3

y=-5/3-1

2x-3y=24

-3y=-2x+24

y=2/3x-8

5 0

3 years ago

Read 2 more answers