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d1i1m1o1n [39]
3 years ago
9

you have a bag with 20 marbles 2 yellow 5 green 12 red and 1 blue which of the following statements are true? Select all that ap

ply. a) The probability of choosing a yellow, green, red, or blue marble is 1 b) The probability of choosing a yellow marble is less than the probability of choosing a blue marble c) The probability of choosing a green marble is 0.25 d) The probability of choosing a black marble is zero
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
5 0
  1. An event has probability 1 if you're certain that it will happen. You are sure to pick a marble that's either yellow, green, red, or blue. So, this probability is 1.
  2. There are more yellow marbles than blue marbles, so it's actually more likely that you'll pick a yellow one, rather than the blue one.
  3. The probability of picking a green marble is given by the number of green marbles divided by the total number of marbles. Compute this ratio and see what you get.
  4. Similarly, the probability of picking a black marble is given by the number of black marbles divided by the total number of marbles. There are no black marbles, so...
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Beth is 5 years younger than Celeste. Next year, their ages will have a sum equal to 57. How old ech now?
RSB [31]

Celeste is 30 and Beth is 25

let x be Celeste's age now then Beth is x - 5

next year they will be 1 year older

Celeste's age will be x + 1 and Beth will be x - 5 + 1 = x - 4

the sum of their ages is therefore

x + 1 + x - 4 = 57

2x - 3 = 57 ( add 3 to both sides )

2x = 60 ( divide both sides by 2 )

x = 30

Celeste is 30 and Beth is 30 - 5 = 25


8 0
2 years ago
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r
Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt

If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

Hence,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

Thus,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

and the Taylor theorem with Lagrange remainder is

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Step-by-step explanation:

5 0
3 years ago
A pilot of a helicopter plans to release a bucket of water on a forest fire. The height y in feet of water t seconds after its r
frosja888 [35]
You’re gonna have to use the quadratic formula here and a bit of thinking. We know y is height, and once it touches the ground, it will be 0.

(-b +- sqrt b^2 - 4ac)/2a
2 +- sqrt(4 + 48(400))/-32
(2 +- 138.58)/-32
We don’t want a negative time because it’ll make no sense. So do subtraction

(2-138.58)/-32 = 4.27 seconds aka 4.3 seconds
4 0
3 years ago
The Mayors of two small towns that adjoin a large community have engaged in an argument on which of their cities is safer, with
Lelu [443]

Answer:

1. City A: 150 violent crimes per 100,000 residents.

   City B: 161 violent crimes per 100,000 residents.

2. City A

Step-by-step explanation:

City A:

Population = 123,000

Violent crimes = 185

Violent crimes per 100,000 people:

V_{100k} = \frac{185*100,000}{123,000} \\V_{100k} = 150

City B:

Population = 84,000

Violent crimes = 135

Violent crimes per 100,000 people:

V_{100k} = \frac{135*100,000}{84,000} \\V_{100k} = 161

1. City A: 150 violent crimes per 100,000 residents.

   City B: 161 violent crimes per 100,000 residents.

2. Even though it has a bigger number of crimes, City A has the lowest crime rate since its population is larger.

5 0
3 years ago
If the probability of selecting, without replacement, 2 red marbles from a bag containing only red and blue marbles is 3 55 and
Marat540 [252]

Answer:

Step-by-step explanation:

Let T = the TOTAL number of marbles in the bag.

We are told that 3 of those marbles are red

P(select a red marble first) = 3/T

Once we've removed the first red marble, there are T-1 marbles remaining and 2 of them are red

So, P(select a red marble second) = 2/(T-1)

Okay, now let's use probability rules to answer the question....

P(select 2 red marbles) = P(select a red marble first AND select a red marble second)

= P(select a red marble first) x P(select a red marble second)

= 3/T x 2/(T-1)

= 6/(T² - T)

We're told that P(select 2 red marbles) = 3/55

So, we can write: 6/(T² - T) = 3/55

Cross multiply to get: 3(T² - T) = (6)(55)

Divide both sides by 3 to get: T² - T = (2)(55)

Evaluate: T² - T = 110

Rearrange: T² - T - 110 = 0

Factor: (T - 11)(T + 10) = 0

So, EITHER T = 11 OR T = -10

Since T cannot be negative, it must be the case that T = 11

6 0
2 years ago
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