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3 years ago

PLEASE HELP ME WITH THIS. THANK U.

Mathematics

2 answers:

Lera25 [3.4K]3 years ago

3 0

NAJAJJHSJAHAKAKLAjJajaa

Marina86 [1]3 years ago

3 0

Answer:

B' is (2, 0) and C' is (7, 1)

Step-by-step explanation:

Compare A and A' and you see that you added 2 to X and 5 to Y. Now you simply apply this to B and C to find B' and C'

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Please Help! Will give BRAINLIEST! I don't know how to do this.

OverLord2011 [107]

Therefore x intercept(a) is <span>3 </span> and y intercept(b) is 3<span> </span>.

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3 years ago

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What is the area of the trapezoid? Leave the answer in simplest radical form. The figure is not drawn to scale.

inn [45]

Answer:

• 72 ft²

Step-by-step explanation:

The bottom side length of the triangle is the difference between the top base length and the bottom base length:

13 ft - 5 ft = 8 ft

The vertical side length of the triangle is the same, because the triangle is an isosceles right triangle.

Since we know the base lengths and height of the trapezoid, we can use the usual formula for the area:

A = (1/2)(b1 +b2)h

= (1/2)(5 ft + 13 ft)(8 ft) = 72 ft²

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3 years ago

In how many ways can 4 light bulbs be selected from a batch of 43 lights bulbs to be tested for defects? Assume no light bulb wi

lesantik [10]

Answer:

There are 123,410 ways

Step-by-step explanation:

The number of ways or combinations in which we can select k elements from a group of n elements is given by:

$\frac{n!}{k!(n-k)!}$

In this case, we can replace n by 43 and k by 4, because we are going to select 4 light bulbs from a batch of 43 light bulbs.

So, the number of combinations is calculated as:

$\frac{43!}{4!(43-4)!}=123,140$

It means that there 123,410 ways to select 4 light bulbs from a batch of 43 lights bulbs to be tested for defects

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3 years ago

If a game consists of picking a number from a bag of beans numbered 2, 2, 3, 4, and 5 and flipping a coin, which probabilities a

Semenov [28]

The true statement is (d) none of the above

<h3>Missing information</h3>

I. P(3 and tail)

II. P(even and head)

III .P(odd and head)

<h3>How to determine the equivalent probabilities?</h3>

We start by calculating each probability

There is only one 3 in the 5 numbers.

So:

P(3) = 1/5

The probability of a tail in a coin is:

P(Coin) = 1/2

So, we have:

P(3 and tail) = 1/5 * 1/2

P(3 and tail) = 1/10

There are three even numbers in the 5 numbers.

So:

P(even) = 3/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(even and head) = 3/5 * 1/2

P(even and head) = 3/10

There are two odd numbers in the 5 numbers.

So:

P(odd) = 2/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(odd and head) = 2/5 * 1/2

P(even and head) = 1/5

By comparing the results, none of the probabilities are equal.

Hence, the true statement is (d) none of the above