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3 years ago

15

Can someone help please?

1 answer:

d1i1m1o1n [39]3 years ago

7 0

wow hows college going any tips for me when i go

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What is the positive solution to the equation 0 = -x2 + 2x + 1?

katrin2010 [14]

Answer:

there is no solution

Step-by-step explanation:

1st switch sides: -x2+2x+1=0

2nd add similar elements: -2x + 2x = 0

1 = 0

Thus, this equation has no solution

hope i helped :)

5 0

4 years ago

Read 2 more answers

PLEASE HELP i will give brainliest no links

frutty [35]

Answer:

y=5x

Step-by-step explanation:

y=ax

use any value of x and y from the table

I chose (2,10)

10=2a

10/2=a

a=5

5 0

3 years ago

The area of the region in the first quadrant between the graph of y = k*sin(x) and x-axis (where k > 0 ) on the closed interv

astraxan [27]

Note that $0\le k\sin x\le k$ in the given closed interval, so the area is exactly given by the definite integral

$\displaystyle\int_0^\pi k\sin x\,\mathrm dx$

(no absolute values needed!)

Integrating gives

$-k\cos x\bigg|_{x=0}^{x=\pi}=-k(\cos\pi-\cos0)=-k(-1-1)=2k$

8 0

4 years ago

What is the length of a missing side of a 48 feet and 52 feet 90° triangle

RideAnS [48]

Answer:

5

Step-by-step explanation:

5 0

3 years ago

100 random students are surveyed outside of the student center on campus. Let V denote that the student has a Visa credit card a

mote1985 [20]

Answer:

(a) The value of P (M | V) is 0.30.

(b) The value of $P (M^{c} | V)$ is 0.70.

(c) The value of P (V | M) is 0.375.

(d) The value of $P(V^{c}|M)$ is 0.625.

Step-by-step explanation:

It is provided that,

<em>V</em> = a student has a Visa card

<em>M</em> = a student has a Master card

N = 100, n (<em>V</em>) = 40, n (<em>M</em>) = 32 and n (<em>V</em> ∩ <em>M</em>) = 12.

The probability of a student having visa card is:

$P(V) = \frac{n(V)}{N}= \frac{40}{100}=0.40$

The probability of a student having master card is:

$P(M) = \frac{n(M)}{N}= \frac{32}{100}=0.32$

The probability of a student having visa card and a master card is:

$P(V\cap M) = \frac{n(V\cap M)}{N}= \frac{12}{100}=0.12$

The conditional probability of an event, say A, given that another event, say B, has already occurred is,

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

(a)

Compute the probability that a student has a master card given that he/she has a visa card also, i.e. P (M | V) as follows:

$P(M|V)=\frac{P(V\cap M)}{P(V)} =\frac{0.12}{0.40}=0.30$

Thus, the value of P (M | V) is 0.30.

(b)

Compute the probability that a student does not have a master card given that he/she has a visa card also, i.e. $P (M^{c} | V)$ as follows:

$P (M^{c} | V)=1-P(M|V)=1-0.30=0.70$

Thus, the value of $P (M^{c} | V)$ is 0.70.

(c)

Compute the probability that a student has a visa card given that he/she has a master card also, i.e. P (V | M) as follows:

$P(V|M)=\frac{P(V\cap M)}{P(M)} =\frac{0.12}{0.32}=0.375$

Thus, the value of P (V | M) is 0.375.

(d)

Compute the probability that a student does not have a visa card given that he/she has a master card also, i.e. $P(V^{c}|M)$ as follows:

$P(V^{c}|M)=1-P(V|M)=1-0.375=0.625$

Thus, the value of $P(V^{c}|M)$ is 0.625.

8 0

3 years ago