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3 years ago

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Mathematics

1 answer:

Korolek [52]3 years ago

7 0

Answer:

Step-by-step explanation:

so I can't see the picture or what the "Shaded region " is.. but I can guess that it's the part of the 3x3 rectangle with the circle removed.. right?

if that's the case... then

3*3 = 9 $cm^{2}$

9 - area of circle = shaded region

area of circle = pi * $r^{2}$

pi * 1 cm = 3.1415 $cm^{2}$

9 $cm^{2}$ - 3.1415 $cm^{2}$ = 5.8585 $cm^{2}$

5.86 $cm^{2}$

I don't see this as a choice of answers so maybe that's not what the shaded region is. any way.. hope it helps

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Let f (x)=x-2 and g(x)=x to the second power -7x-9.find f (g(-1))

Degger [83]

f(g(-1)) = - 3

Evaluate g(-1) and substitute into f(x)

g(-1) = (-1)² -7(-1) - 9 = 1 + 7 - 9 = - 1

f(g(-1)) = (-1) - 2 = - 3

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3 years ago

A TV is listed for 35% off. If the original price was $550, what is the sale price?

almond37 [142]

Answer:

357.50$

Step-by-step explanation:

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3 years ago

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In a large population, about 10% of people do not like the taste of cilantro, an herb used in cooking. A researcher takes a rand

bogdanovich [222]

Answer:

0.0019390

Step-by-step explanation:

10% of people do not like the taste of cilantro

A researcher takes a random sample of 15 people and surveys whether they like cilantro

Use the binomial distribution to compute the probability that exactly 6 of the people in the sample do not like cilantro.

Probability of success = 0.10 = p

Probability of failure = 0.9 =q

n = 15

r = 6

Formula : $P(X=r)=^nC_rp^r q^{n-r}$

$P(X=6)=^{15}C_6 (0.1)^6 (0.9)^{15-6}$

$P(X=6)=\frac{15!}{6!(15-6)!} (0.1)^6 (0.9)^{15-6}$

$P(X=6)=0.0019390$

Hence the probability that exactly 6 of the people in the sample do not like cilantro is 0.0019390

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3 years ago

Describe the steps you can use to find an average of fractional amounts

lutik1710 [3]

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3 years ago

Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or impo

Rudik [331]

Answer:

a) Probability that vehicle is Light Truck is 0.69

b) Probability that vehicle is imported car is 0.08 or 8%.

Step-by-step explanation:

Part a)

Event that vehicle is a car is given by A. Since, the vehicles can only be classified as cars or light trucks, the event that a vehicle will be light truck will be compliment of A.

i.e. Event that vehicle is a Light Truck = $A^{c}$

It is given that in recent year 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.

So, from here we can say that, if a vehicle is randomly selected from all the vehicles the probability that it would be a light truck will be:

P(Vehicle will be Light Truck) = P( $A^{c}$ ) = 69% = 0.69

Part b)

Event that vehicle is imported is given by B. We need to find the probability the a randomly chosen vehicle is an imported car i.e. we have to find probability of occurrence of events A and B together, which will be denoted as: P(A and B)

Since, P( $A^{c}$ ) = 69% = 0.69

P(A) = 1 - P( $A^{c}$ ) = 1 - 0.69 = 0.31

It is also given that 78% of vehicles were domestic. This means, the percentage/probability of imported vehicles is:

P(B) = 1 - 0.78 = 0.22

55% of vehicles were domestic light trucks. This can be expressed as:

$P(A^{c}\&B^{c})$ , the compliment of this event will give us P(A or B).

i.e.

P(A or B) = 1 - $P(A^{c} \&B^{c})$ = 1 - 0.55 = 0.45

According to the addition rule of probabilities:

P(A or B) = P(A) + P(B) - P(A and B)

Substituting the calculated values gives us:

0.45 = 0.31 + 0.22 - P(A and B)

P(A and B) = 0.08

This means, the probability that vehicle is imported car is 0.08 or 8%.

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