Ask question

Ask question

All categories

2 years ago

8

On Wednesday, Danielle skated 2/3 of the way around the track in 2 minutes. Her younger brother skated 3/4 of Danielle's distanc

e in 2 minutes. What fraction of the track did Danielle's brother finish in 2 minutes?

1 answer:

katen-ka-za [31]2 years ago

4 0

75 percent of 2/3

Is 0.5 of the way

You might be interested in

In triangle ABC segment DE is parallel to the side AC. (The endpoints of segment DE lie on the sides AB and BC respectively).

ivanzaharov [21]

Answer:

6 dm

Step-by-step explanation:

Triangle DBE is similar to triangle ABC, so their side lengths are proportional.

DE/AC = DB/AB

The length of DB can be found from ...

DB +AD = AB

DB = AB -AD = (15 -10) dm = 5 dm

So, we can fill in the proportion:

DE/(18 dm) = (5 dm)/(15 dm)

DE = (18 dm)·(1/3) . . . . . . . . . . simplify, multiply by 18 dm

DE = 6 dm

_____

It can be helpful to draw and label a figure.

5 0

1 year ago

A preimage includes a line segment with a length of x units and a slope of m units. The preimage is dilated by a scale factor of

lapo4ka [179]

Nx is the first one and m is the second answer

8 0

2 years ago

Read 2 more answers

The production department has installed a new spray machine to paint automobile doors. As is common with most spray guns, unsigh

Nesterboy [21]

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter $\lambda =0.5$ / door

The probability mass function on of a poisson distribution Is:

$P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}$

$P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}$

The probability that no blemishes occur is :

$P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}$

$P(X=0) = 0.60653$

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes is = qp

=10,000(0.6065)

= 6065

3 0

2 years ago

In high-school 135 freshmen were interviewed.

timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

3 0

1 year ago

Read 2 more answers

Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .

lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

b) $P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this using independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

If we define the random of variable Y we know that:

$Y\sim Geo (1-p)$

Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

8 0

2 years ago