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2 years ago

PLEASE HELP DUE IN 15 MINS!!!!!!!!!!!!

Mathematics

1 answer:

Alborosie2 years ago

4 0

$∠3 =∠4$

$∠S +∠T=90°$

$∠X+∠Y=180°$

$∠2+∠4=90°$

$∠J+∠K = 180°$

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Please help me I will give you the brai thing and extra points, it's math

Lesechka [4]

Answer:

Step-by-step explanation:

6 0

2 years ago

Earth is represented on a map of a portion of the solar system so that its surface is the circle with equation x2+y2+8x+2y−4883=

vodka [1.7K]

Answer:

The general form of the equation for the orbit of the satellite is $x^{2}+y^{2}+8\cdot x+2\cdot y = 4939.16$ .

Step-by-step explanation:

From statement we have the general equation of the elliptic form of Earth. First we have to transform this formula to the standard form of the equation of the circle. That is:

1) $x^{2}+y^{2}+8\cdot x +2\cdot y -4883=0$ Given

2) $(x^{2}+8\cdot x)+(y^{2}+2\cdot y) = 4883$ Associative, commutative and modulative properties/Compatibility with addition/Existence of additive inverse.

3) $(x^{2}+8\cdot x +16)+(y^{2}+2\cdot y+1)-17=4883$ Associative, commutative and modulative properties/Compatibility with addition/Existence of additive inverse.

4) $(x+4)^{2}+(y+1)^{2} = 4900$ Square perfect trinomial/Result

According the result, the center of Earth is $C(x,y) = (4, 1)$ and the square of the radius of Earth equals 4900. That is: $r = 70$

If satellite circles 0.4 unit above Earth, then the equation of the orbit of the satellite is:

$(x+4)^{2}+(y+1)^{2}=70.4^{2}$

$(x+4)^{2}+(y+1)^{2} = 4956.16$

Now we transform this into its general form:

1) $(x+4)^{2}+(y+1)^{2} = 4956.16$ Given

2) $x^{2}+8\cdot x +16+y^{2}+2\cdot y+1 = 4956.16$ Square perfect trinomial

3) $x^{2}+y^{2}+8\cdot x+2\cdot y = 4939.16$ Associative, commutative and modulative properties/Compatibility with addition/Existence of additive inverse/Result

The general form of the equation for the orbit of the satellite is $x^{2}+y^{2}+8\cdot x+2\cdot y = 4939.16$ .

6 0

3 years ago

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic

stellarik [79]

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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 problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

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

$P(X = 0) = C_{10,0}*(0.05)^{0}*(0.95)^{10} = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

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

$P(X = 1) = C_{10,1}*(0.05)^{1}*(0.95)^{9} = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_{10,2}*(0.05)^{2}*(0.95)^{8} = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_{10,0}*(0.95)^{0}*(0.05)^{10}\cong 0$

$P(X = 1) = C_{10,1}*(0.95)^{1}*(0.05)^{9} \cong 0$

$P(X = 2) = C_{10,1}*(0.95)^{2}*(0.05)^{8} \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

8 0

3 years ago

Can anyone help me please solving these maths problems, first time. Thanks.

olga_2 [115]

I think the answer is 2/2

5 0

3 years ago

Ada had $42 to spend on a binder and 11 spiral notebooks. The binder cost $9 what was the price of each spiral notebook?

ArbitrLikvidat [17]

The price for each one was 3 dollars :)
9 + 11x = 42
subtract 9 from each side
11x = 33
divide by 11
x = 3

4 0

3 years ago

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PLEASE HELP DUE IN 15 MINS!!!!!!!!!!!!​

PLEASE HELP DUE IN 15 MINS!!!!!!!!!!!!