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3 years ago

Plsss freakin help me :((

Mathematics

2 answers:

Harrizon [31]3 years ago

3 0

Answer:

D, last one

Step-by-step explanation:

marusya05 [52]3 years ago

3 0

Answer:

The choice four;

$w = \frac{24}{lh}$

Step-by-step explanation:

$24 = lwh \\ \\ 24 = (lh) w \\ \\ w = \frac{24}{lh}$

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800 is 100 times as much as

Slav-nsk [51]

800/100 = 8
800 is 100 times as much as 8

8 0

3 years ago

X − 6 = 63 − 10(x + 8)

il63 [147K]

$\red { \green {\boxed {\boxed{Answer}}}}$

$x - 6 =63 - 10(x + 8)$

$x - 6 = 63 - 10x - 80$

$x - 6 = 63 - 80 - 10x$

$x - 6 = - 17 - 10x$

$x - 6 + 10x = - 17$

$x + 10 x- 6 = - 17$

$11x - 6 = - 17$

$11x = - 17 + 6$

$11x = - 11$

$x = \frac{ - 11}{11}$

$x = - 1$

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8 0

3 years ago

Read 2 more answers

Sketch the area under the standard normal curve over the indicated interval and find the specified area. (You can use the probab

mafiozo [28]

The area to the right of z = 1.35 is 0.0885 and the area to the left of -0.47 is 0.3192.

<h3>How to compute the values?.</h3>

Given z = 1.35

= 1- P(z < 1.35)

= 1- 0.9115

= 0.0885

The area to the left of -0.47 will be:

= 1 - P(z < 0.47)

= 1 - 0.6808

= 0.3192

Learn more about normal curve on:

brainly.com/question/6758792

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8 0

2 years ago

A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

Oksana_A [137]

Answer:

Step-by-step explanation:

Given that a bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7.

From the above we cannot conclude that red cards and even numbers are mutually exclusive

Just drawing two red cards and because the two happen to be odd we cannot generalize the red cards have odd numbers.

This might have occurred due to simple chance from a comparatively large number of 40 cards.

Suppose say we have red cards 20, and 19 red 1 blue.

Then drawing 2 from 19 red cards have more probability and this can occur by chance.

So friend's conclusion is wrong.

5 0

3 years ago

Figure ABCD is graphed on a coordinate plane.

Phoenix [80]

1. BC is 1+3=4 (units)

2. AD is 3+5= 8 (units)

3. Draw the altitude CH from point C, as shown in the figure.

4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.

5. $CD^2=HD^2+CH^2$

$CD^2=2^2+4^2=4+16=20$

$CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} * \sqrt{5}=2 \sqrt{5}= 4.47$

6. AB=CD=4.47 because the trapezoid is isosceles.

7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)

3 0

4 years ago

Read 2 more answers