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3 years ago

Helppppppppp please

Mathematics

2 answers:

Sladkaya [172]3 years ago

6 0

Answer:

44m^2

I am attaching the work below.

mariarad [96]3 years ago

4 0

Well here's the equation 4x3 x2=24 for both of the sides and the top 10x2 because if 4 is the bottom and the sides are 6 6-4=2 so 10x2=20 and we add the areas and

20

+

24

_____

44 is the combined areas (Your answer 44)

HOPE THIS HELPS :D

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Pleasse and thank u <33

liubo4ka [24]

Answer:

4x-2x+3y+6x+6y distributive property

4x-2x+6y+3y+6y commutative property of addition

8x+9y combine like terms

Step-by-step explanation:

I hope this helps!

7 0

2 years ago

(-4,19) (0,3) whats the slope?

Troyanec [42]

The slope is 16/4 which is 4/1 which can also be 4. So the most simplified is 4

4 0

2 years ago

Read 2 more answers

Assume the random variable x has a binomial distribution with the given probability of obtaining a success. Find the following.

borishaifa [10]

Answer:

P(x > 10) = 0.6981.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question:

$n = 14, p = 0.8$

P(x>10)

$P(x > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 11) = C_{14,11}.(0.8)^{11}.(0.2)^{3} = 0.2501$

$P(X = 12) = C_{14,12}.(0.8)^{12}.(0.2)^{2} = 0.2501$

$P(X = 13) = C_{14,13}.(0.8)^{13}.(0.2)^{1} = 0.1539$

$P(X = 14) = C_{14,14}.(0.8)^{14}.(0.2)^{0} = 0.0440$

$P(x > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.2501 + 0.2501 + 0.1539 + 0.0440 = 0.6981$

So P(x > 10) = 0.6981.

8 0

3 years ago

The population of a town gradually decreases by about 30

topjm [15]

It would be 3,600 - 30 = 3570

3 0

2 years ago

The American Veterinary Association claims that the annual cost of medical care for dogs averages $100 with a standard deviation

Taya2010 [7]

Answer:

The probability that the cost for someone's dog is higher than for the cat is 33.2%.

Step-by-step explanation:

With the data given, we know that the difference in cost of medical care of dogs and cats have a normal distribution with μ=-20 and σ=46.

To know the probability of the cost for the dog is higher than the cat, we have to calculate the probability of P(d>0).

Then we have to calculate z, and look up in a standarized normal distribution table.

Calculate z:

$z=\frac{X-\mu}{\sigma}=\frac{0-(-20)}{46}=\frac{20}{46}= 0.4348$

The probability of the difference being higher than 0, we have:

$P(d>0)=P(z>0.4348)=0.33185$

We can say that the probability that the cost for someone's dog is higher than for the cat is 33.2%.

8 0

2 years ago