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2 years ago

PLS ANSWER PLS IM DESPRATE WILL MARK BRAINLIEST 18 POINTS

Mathematics

1 answer:

Jlenok [28]2 years ago

6 0

D I believe because if you look at point B it is (3,3) but dilated to (9,9) and 9/3=3. Which also means that the side lengths will be 3 times the original triangle’s side lengths

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Can someone solve this please 9g = 36

Goryan [66]

You divide 9 on both sides than it would be g=4

3 0

3 years ago

Read 2 more answers

Which is the equation of an ellipse centered at the origin with foci on x-axis, x-intercepts +7, -7 and y-intercepts +2,

Aleks04 [339]

We know that
foci on x-axis-------> is a ellipse with horizontal major axis

the equation is
(x²/a²)+(y²/b²)=1

major axis is the x-axis with length 2a
2a=14--------> a=7

minor axis is the y-axis with length 2b
2b=4-----> b=2

the equation is
(x²/a²)+(y²/b²)=1------> (x²/7²)+(y²/2²)=1-----> (x²/49)+(y²/4)=1

the answer is
(x²/49)+(y²/4)=1

see the attached figure

7 0

3 years ago

The random variable X~(30,2^2) Find p(X<33) Find p(X>26)

Ostrovityanka [42]

Answer:

i) P(X<33) = 0.9232

ii) P(X>26) = 0.001

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 30

Given that the standard deviation of the Population = 4

Let 'X' be the Normal distribution

Step(ii):-

Given that the random variable X = 33

$Z = \frac{x-mean}{S.D}$

$Z = \frac{33-30}{2} = 1.5$ >0

P(X<33) = P( Z<1.5)

= 1- P(Z>1.5)

= 1 - ( 0.5 - A(1.5))

= 0.5 + 0.4232

P(X<33) = 0.9232

Step(iii) :-

Given that the random variable X = 26

$Z = \frac{x-mean}{S.D}$

$Z = \frac{33-26}{2} = 3.5$ >0

P(X>26) = P( Z>3.5)

= 0.5 - A(3.5)

= 0.5 - 0.4990

= 0.001

P(X>26) = 0.001

5 0

2 years ago

Consider the following information related to a set of quantitative sample data:

svlad2 [7]

Answer:

(a) There are outliers

(b) $x$ and $x >62$

Step-by-step explanation:

Given

$\sigma = 14.92$

$\bar x = 22.0$

$q_0 = -24$

$q_1 = 14.5$

$q_2 = 24.5$

$q_3 = 33.5$

$q_4 = 64$

Solving (a): Check for outliers

This is calculated using:

$Lower = Q_1 - (1.5 * IQR)$ --- lower bound of outlier

$Upper = Q_3 +(1.5 * IQR)$ --- upper bound of outlier

Where

$IQR = Q_3 - Q_1$

So, we have:

$IQR = 33.5 - 14.5$

$IQR = 19$

The lower bound of outlier becomes

$Lower = Q_1 - (1.5 * IQR)$

$Lower = 14.5 - (1.5 * 19)$

$Lower = 14.5 - 28.5$

$Lower = -14$

The upper bound of outlier becomes

$Upper = Q_3 +(1.5 * IQR)$

$Upper = 33.5 + 1.5 * 19$

$Upper = 33.5 + 28.5$

$Upper = 62$

So, we have:

$-14 \le x \le 62$ --- the range without outlier

Given that:

$q_0 = -24$ --- This represents the lowest data

$q_4 = 64$ --- This represents the highest data

-24 and 64 are out of range of $-14 \le x \le 62$ .

Hence, there are outliers

Solving (b): The outliers

The outliers are data less than the lower bound (i.e. less than -14) or greater than the upper bound (i.e. 62)

So, the outliers are:

$x$ and $x >62$

4 0

2 years ago

Using the given information write the equation of parabolas:

lana [24]

The general form of the equation we need to find is (x - h)^2 = 4p(y- k).

The center is the distance between the directrix and focus.

So, center (h, k) = (3, 3/2) .

P = distance from center to the focus and it just so happens to be 1.5.

We now plug everything into the formula given above.

(x - 3)^2 = 4(1.5)(y - 3/2)

(x - 3)^2 = 6(y - 3/2)

Done!

8 0

3 years ago