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3 years ago

Lines l and m are parallel. Which angle is a corresponding angle with ∠5? ∠7 ∠6 ∠8

Mathematics

1 answer:

Amanda [17]3 years ago

5 0

Answer:

∠7

Step-by-step explanation:

Given that lines l and m are parallel, and line t (the transversal) runs across both lines, thus:

Angle 5 and angle 7 occupy the same relative position or corner. They have the same matching corners where the line intercepts. Therefore, both angles are said to be corresponding angles.

∠7 is a corresponding angle with ∠5. ∠7 = ∠5.

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he mean incubation time of fertilized eggs is 23 days. Suppose the incubation times are approximately normally distributed with

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I think a but I’m not quite sure

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3 years ago

Write an equation of the line that passes through (2, -5) and is parallel to the line 2y = 3x + 10

Zarrin [17]

First you get "y" by itself. To do so you divide 2 on both sides.

y = 3/2x + 5

To write an equation of a line that is PARALLEL to this equation, the slopes have to be the SAME. So the slope is 3/2.

You then use the equation:

y = mx + b

SInce you know "m" you plug it in.

y = 3/2x + b

Now you need to find b. To do so you plug in the point (2, -5) into this equation.

-5 = 3/2(2) + b

-5 = 3 + b

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Finally you plug in b and you get your new equation.

y = 3/2x - 8

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3 years ago

A population of bacteria is introduced into a culture. the number of bacteria P can be modeled by P=500(1+4t/(50+t^2 )) where t

Dennis_Churaev [7]

P(t)=500(1+4t/(50+t^2 ))

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by the quotient rule

500 (-4t^2 + 200)/(t^2 + 50)^2

Hence

P'(2) = 500 . (-16 + 200)/54^2 ~= 31.6

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4 years ago

9.50 One way to evaluate the effectiveness of a teaching assistant is to examine the scores achieved by his or her students on a

EleoNora [17]

Answer:

Step-by-step explanation:

Given that a professor sets a standard examination at the end of each semester for all sections of a course. The variance of the scores on this test is typically very close to 300.

$H_0: s^2 = 300\\H_a: s^2 \neq 300$

(Two tailed test for variance )

Sample variance =480

We can use chi square test for testing of hypothesis

Test statistic = $\frac{(n-1)s^2}{\sigma^2} \\=62.4$

p value = 0.0100

Since p <0.05 our significance level, we reject H0.

The sample variance cannot be claimed as equal to 300.

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3 years ago

Due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed into a 18-pack. Suppos

crimeas [40]

Answer:

a) There is a 1.21% probability that both contain diet soda.

b) There is a 79.21% probability that both contain diet soda.

c) $P(X = 2)$ is unusual, $P(X = 0)$ is not unusual

d) There is a 19.58% probability that exactly one is diet and exactly one is regular.

Step-by-step explanation:

There are only two possible outcomes. Either the can has diet soda, or it hasn't. So we use the binomial probability distribution.

Binomial probability distribution

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

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

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 $\pi$ is the probability of X happening.

A number of sucesses $x$ is considered unusually low if $P(X \leq x) \leq 0.05$ and unusually high if $P(X \geq x) \geq 0.05$

In this problem, we have that:

Two cans are randomly chosen, so $n = 2$

Two out of 18 cans are filled with diet coke, so $\pi = \frac{2}{18} = 0.11$

a) Determine the probability that both contain diet soda. P(both diet soda)

That is P(X = 2).

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

$P(X = 2) = C_{2,2}(0.11)^{2}(0.89)^{0} = 0.0121$

There is a 1.21% probability that both contain diet soda.

b)Determine the probability that both contain regular soda. P(both regular)

That is P(X = 0).

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

$P(X = 0) = C_{2,0}(0.11)^{0}(0.89)^{2} = 0.7921$

There is a 79.21% probability that both contain diet soda.

c) Would this be unusual?

We have that $P(X = 2)$ is unusual, since $P(X \geq 2) = P(X = 2) = 0.0121 \leq 0.05$

For P(X = 0), it is not unusually high nor unusually low.

d) Determine the probability that exactly one is diet and exactly one is regular. P(one diet and one regular)

That is P(X = 1).

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

$P(X = 1) = C_{2,1}(0.11)^{1}(0.89)^{1} = 0.1958$

There is a 19.58% probability that exactly one is diet and exactly one is regular.

8 0

3 years ago