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2 years ago

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Ms. Turner drove 825 miles in March. She drove 3 times as many miles in March as she did in January. She drove 4 times as many m

iles in February as she did in January. What was the total number of miles Ms. Turner drove in February?

1 answer:

navik [9.2K]2 years ago

4 0

825/3 = 275 which is how many miles she drove in January.

275*4= 1100 which is how many miles she drove in February.

Ms. Turner drove 1100 miles in February.

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If Angle 6 is congruent to angle 10 and Angle 5 is congruent to angle 7, which describes all the lines that must be parallel? Li

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Answer:

Third option: Lines "r" and "s" and lines "t" and "u" must be parallel.

Step-by-step explanation:

The missing figure is attached.

You need to remember that:

1- A Transversal is defined as a line that intersects two or more lines.

2- When a transversal cut two parallel lines, several angles are formed, which are grouped in pairs. Some of them are:

a. Vertical angles: are those pairs of angles that share the same vertex and are opposite to each other. These angles are congruent.

b. Corresponding angles: are those pairs of non-adjacent angles located on the same side of the transversal and outside the parallel lines. They are congruent.

In this case, you can identify in the figure that:

$\angle 6$ and $\angle 10$ are Corresponding angles.

$\angle5$ and $\angle 7$ are Vertical angles.

Therefore, based on the explained before, you can conclude that lines "r" and "s" and lines "t" and "u", must be parallel.

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1 year ago

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The chance she will pull out a red paper is 20%.

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1 year ago

The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

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Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

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Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

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

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

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1 year ago

The vertices of rhombus defg are d(1, 4), e(4, 0), f(1, –4), and g(–2, 0). what is the perimeter of the rhombus?

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Answer: 20 unit.

Step-by-step explanation:

Since, Here the vertices of the rhombus defg are d(1, 4), e(4, 0), f(1, –4), and g(–2, 0).

Where, de, ef, fg, gd are sides of the rhombus defg.

By the distance formula,

$de = \sqrt{(4-1)^2+(0-4)^2}$

$=\sqrt{3^2+4^2}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$= 5$

Thus, the side of rhombus = 5

By the property of rhombus,

de = ef = fg = gd = 5 unit.

Thus, the perimeter of the given rhombus defg = de + ef + fg + gd = 5+5+5+5 = 20 unit

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1 year ago

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Answer:

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Step-by-step explanation:

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I am joyous to assist you anytime.

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2 years ago

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