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ch4aika [34]
3 years ago
8

ABCD is a rectangle. Its diagonals are AC and BD. If AC = 12cm, then BD = _______cm

Mathematics
1 answer:
defon3 years ago
8 0

Answer:

BD = 12\ cm

Step-by-step explanation:

Given

Shape: Rectangle

Diagonals: AC and BD

AC = 12\ cm

Required

Determine BD

From the question, we understand that the shape is a rectangle and its diagonals are AC and BD

Every rectangle has 2 diagonals, both of which are equal.

This implies that:

BD = AC

Recall that:

AC = 12\ cm

Hence,

BD = 12\ cm

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Solve for x. 6x−2/4=x−2 Enter your answer in the box. x =
Goryan [66]

6x−2/4=x−2

⇒ 6x−1/2=x−2

⇒6x−x=1/2−2

⇒5x=−3/2

⇒x=−3/10

Hopes it will hwlp you.

6 0
3 years ago
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What is the equation for the hyperbola shown?
castortr0y [4]

Answer:

\frac{x^{2} }{3600} - \frac{y^{2} }{121} = 1

Step-by-step explanation:

Since the hyperbola opens left and right with a center at the origin, then

\frac{x^{2} }{^{2} } - \frac{y^{2} }{b^{2} } = 1

In the graph a = 60 and b = 11

So

\frac{x^{2} }{60^{2} } - \frac{y^{2} }{11^{2} } = 1.

\frac{x^{2} }{3600} - \frac{y^{2} }{121} = 1

7 0
3 years ago
) All human blood can be typed as one of O, A, B, or AB. The distribution of the type varies a bit with race. For African-Americ
ivann1987 [24]

Answer:

The correct option is 1 - [(0.8)¹⁰+10*0.2*(0.8)⁹]= 0.6242

Step-by-step explanation:

Hello!

Given the distribution of probabilities for blood types for African-Americans:

O: 0.4

A: 0.2

B: 0.32

AB: 0.08

A random sample of 10 African-American is chosen, what is the probability that 2 or more of them have Type A blood?

Let X represent "Number of African-Americans with Type A blood in a sample of 10.

Then you have two possible outcomes,

"Success" the person selected has Type A blood, with an associated probability p= 0.2

"Failure" the selected person doesn't have Type A blood, with an associated probability q= 0.8

(You can calculate it as "1-p" or adding all associated probabilities of the remaining blood types: 0.4+0.32+0.08)

Considering, that there is a fixed number of trials n=10, with only two possible outcomes: success and failure. Each experimental unit is independent of the rest and the probability of success remains constant p=0.2, you can say that this variable has a Binomial distribution:

X~Bi(n;p)

You can symbolize the asked probability as:

P(X≥2)

This expression includes the probabilities: X=2, X=3, X=4, X=5, X=6, X=7, X=8, X=9, X=10

And it's equal to

1 - P(X<2)

Where only the probabilities of X=0 and X=1 are included.

There are two ways of calculating this probability:

1) Using the formula:

P(X)= \frac{n!}{(n-X)!X!} *p^{x} * q^{n-x}

With this formula, you can calculate the point probability for each value of X=x₀ ∀ x₀=1, 2, 3, 4, 5, 6, 7, 8, 9, 10

So to reach the asked probability you can:

a) Calculate all probabilities included in the expression and add them:

P(X≥2)= P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + X=10

b) Use the complement rule and calculate only two probabilities:

1 - P(X<2)= 1 - [P(X=0)+P(X=1)]

2) Using the tables of the binomial distribution.

These tables have the cumulative probabilities listed for n: P(X≤x₀)

Using the number of trials, the probability of success, and the expected value of X you can directly attain the corresponding cumulative probability without making any calculations.

>Since you are allowed to use the complement rule I'll show you how to calculate the probability using the formula:

P(X≥2) = 1 - P(X<2)= 1 - [P(X=0)+P(X=1)] ⇒

P(X=0)= \frac{10!}{(10-)0!0!} *0.2^{0} * 0.8^{10-0}= 0.1074

P(X=1)= \frac{10!}{(10-1)!1!} *0.2^{1} * 0.8^{10-1}= 0.2684

⇒ 1 - (0.1074+0.2684)= 0.6242

*-*

Using the table:

P(X≥2) = 1 - P(X<2)= 1 - P(X≤1)

You look in the corresponding table of n=10 p=0.2 for P(X≤1)= 0.3758

1 - P(X≤1)= 1 - 0.3758= 0.6242

*-*

Full text in attachment.

I hope it helps!

8 0
3 years ago
Plz help ...........
bogdanovich [222]

Step-by-step explanation:

Given c = longest of triange = 15m

Given b = short side of triangle = 9m

By pythagoras' Theorem,

{c}^{2}  =  {a}^{2}   +  {b}^{2}  \\  {a}^{2}  +  {b}^{2} =  {c}^{2}   \\  {a}^{2}  +  {(9)}^{2}  =  {(15)}^{2}  \\  {a}^{2}  + 81 = 225 \\  {a}^{2}  = 225 - 81 \\  {a}^{2}  = 144 \\ a =  \sqrt{144}  \\  = 12m

6 0
3 years ago
If a system of linear equations has no solution, what does this mean about the two lines?​
romanna [79]

Answer:

The lines do not intersect, it is parallel therefore there is no solution.

Step-by-step explanation:

7 0
3 years ago
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