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Stells [14]
2 years ago
5

Lil help here please​

Mathematics
2 answers:
stepan [7]2 years ago
3 0

Answer:

d=-48

Step-by-step explanation:

d=-8x6=-48

-BARSIC- [3]2 years ago
3 0

Answer and Step-by-step explanation:

To solve for d, we need to get it by itself. We do that by multiplying 6 to both sides of the equation.

d = -48

When we plug -48 in for d we get:

\frac{-48}{6}  = -8.

<u>That means the answer is indeed d = -48.</u>

<u></u>

<u></u>

<u></u>

<em><u>#TeamTrees #PAW (Plant And Water)</u></em>

<em><u></u></em>

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A figure is composed of a rectangle and a right triangle. What is the area of the figure?
strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h

We substitute the values to obtain;

Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm

Area=\frac{1}{2}(21.6cm)\times 6cm

Area=64.8cm^2

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

C=2\pi r

where r=3.5ft is the radius of the circular table cloth.

We substitute this value and \pi=3.14 into the formula to get;

C=2(3.14)(3.5)ft

C=21.98ft

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

C=\pi r

where r=12.5cm is the radius of the semicircle.

C=3.14\times 12.5cm

C=39.25cm

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh

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Area=345.3125+300

Area=645.3125cm^2

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QUESTION 6

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A=bh

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We substitute the values to obtain;

23.7=3y

y=\frac{23.7}{3}

y=7.9in.

QUESTION 7

The area of a rectangle is given by the formula;

Area=l\times b

The area of the bigger rectangle =9\times15=135in^2

The area of the smaller rectangle =8\times 13=104in^2

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5 0
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vodomira [7]

Answer:

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Step-by-step explanation:

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi
svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62%  probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

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.

3% of the components are identified as defective

This means that p = 0.03

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is P(X = 0) when n = 90. So

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

P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so P(X \leq 2) when n = 102

Then

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

In which

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

P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447

P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411

P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062

0.4062 = 40.62%  probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so P(X \leq 5) when n = 105

Then

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

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

P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408

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P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133

P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265

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