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Pie
3 years ago
9

Solve for A if Ab = ut Solve for c if Kc = 9 a Solve for D if Bd = Ca

Mathematics
1 answer:
pychu [463]3 years ago
7 0

Answer:

Below in bold.

Step-by-step explanation:

Ab = ut

Divide both sides by b

A = ut/b.

The other 2 are solved in a similar way:

Kc = 9a

c = 9a/K

Bd = Ca

d = Ca/B.

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Julie has the three segments shown in the graph. Can she make a triangle with these segments.
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3 years ago
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PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab
Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

m=\frac{y_2-y_1}{x_2-x_1}

The product of slopes of two perpendicular lines is -1.

Slope of GR is

\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}

Slope of RT is

\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1

Product of slopes of GR and RT is

\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

(x,y)\rightarrow(x-1,y-2)

Then the coordinates of T' are

(2,6)\rightarrow(2-1,6-2)

(2,6)\rightarrow(1,4)

Slope of RT' is

\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}

Product of slopes of GR and RT' is

\frac{-4}{3}\times \frac{3}{4}=-1

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

6 0
3 years ago
A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando
kipiarov [429]

The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

<u>P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ</u>

where n = total no. of trials

          p = probability of success

          x = no. of successful trials

          q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

          = 1 - (0.98)⁴

          = 1 - 0.9224

P(X≥1) = 0.0776

4 0
3 years ago
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