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3 years ago

3(x – 5) = 2(x – 5) + x solve for x and number of soultions

Mathematics

2 answers:

ElenaW [278]3 years ago

5 0

Answer: X=0

Explanation:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

Step-by-step explanation

3x-15=2x-10+x

3x-15=3x-10

in this step, problem happens

becoz both side doesnt equal, no ans can be proven

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How do I solve for h

WITCHER [35]

All the triangles are similar, so the ratio of (long leg)/(short leg) is the same. The long leg of ΔADB is
.. AD = AC -DC = 20 -4
.. AD = 16

Then
.. (long leg)/(short leg) = AD/DB = BD/DC
.. 16/h = h/4
.. h^2 = 64
.. h = 8

Selection D is appropriate.

_____
You may see this again. The altitude (h) from the right angle in a right triangle is the geometric mean of segments AD and DC. That is,
.. h = √((AD)*(DC))

8 0

3 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

3 years ago

17) List the ways five dogs a Beagle, a Collie, a Greyhound, a Poodle and a

Gnoma [55]

Answer:

10 pairs

It is a combination

Explanation:

This is a combination because the order does not matter: the Beagle - Collie couple is the same as the Collie - Beagle couple, so it is only counted once.

The formula for combinations is:

$C(m,n)=\dfrac{m!}{n!\cdot (m-n)!}$

For five diffferent dogs to be put in pairs, the number of different pairs is:

$C(5,2)=\dfrac{5!}{2!\cdot (5-2)!}=\dfrac{5\times 4}{2}=10$

4 0

3 years ago

In what form is the following linear equation written v=9x+2

Oksi-84 [34.3K]

Taking v to be y, it's written in the standard form for a linear equation, y=mx+c

6 0

3 years ago

The length is _____ the width is______ and the area of the actual room is ___ _______

RSB [31]

Answer:

98 square inches.

Step-by-step explanation:

The length is 14 inches. The width is 7 inches.

7 * 14 = 98 square inches.

The area is 98 square inches.

8 0

3 years ago