PLZ Needed NOW plz PLZZZ IT'S URGENT

If I have an equation in which x = pounds, and y = total cost, can I substitute quarts into the equation

Colt1911 [192]

Answer:

Yes, you can

Step-by-step explanation:

As a point of reference, assume the equation is:

$y = 5x$

Where

$y = total\ cost$

and

$x = pounds$

In standard unit of conversion:

$1\ pound = 0.4793\ quart$

So:

$x\ pounds = 0.4793x\ quart$

Substitute 0.4793x for x in $y = 5x$

$y = 5 * 0.4793x$

$y = 2.3965x$

The above equation is the equivalent of $y = 5x$ in quarts

<em>So, irrespective of what the equation is, you can always substitute quarts into the equation.</em>

8 0

3 years ago

An automobile manufacturer has discovered that 20% of all the transmissions it installed in a particular style of truck are defe

Hatshy [7]

Answer:

0.148 = 14.8% probability that they will need to order at least one more new transmission

Step-by-step explanation:

For each transmission, there are only two possible outcomes. Either it is defective after a year of use, or it is not. The probability of a transmission being defective is independent of any other transmission. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all the transmissions it installed in a particular style of truck are defective after a year of use.

This means that $p = 0.2$

Sold seven trucks:

This means that $n = 7$

It has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission?

This is the probability that at least 3 are defective, that is:

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{7,0}.(0.2)^{0}.(0.8)^{7} = 0.2097$

$P(X = 1) = C_{7,1}.(0.2)^{1}.(0.8)^{6} = 0.3670$

$P(X = 2) = C_{7,2}.(0.2)^{2}.(0.8)^{5} = 0.2753$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2097 + 0.3670 + 0.2753 = 0.852$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.852 = 0.148$

0.148 = 14.8% probability that they will need to order at least one more new transmission

6 0

3 years ago

What is the interest on RS. 1000 at 10 p.c.p.a for 1 year.

solniwko [45]

Answer:

Interest wil be 10% of 100 which is 10

So, interest will be Rs.10

And total amount to be payed will be 100+10 which is Rs.110

8 0

3 years ago

Read 2 more answers

Without using a calculator, what is 30% of 546? (You can round up to avoid decimal points)

madreJ [45]

Answer:

Hey there!

We can solve this by multiplying 0.3 by 546, which is about 164.

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Please help me !!!!!!

DIA [1.3K]

Answer:

(x, y) = (2, 2)

Step-by-step explanation:

The graph is attached.

Both equations are in slope-intercept form:

y = mx +b . . . . . . line with slope m and y-intercept b

The graph of the first equation intersects the y-axis at +3, and has a slope (rise/run) of -1/2. That is, it decreases 1 unit for each 2 units to the right.

The graph of the second equation intersects the y-axis at -4, and has a slope of +3. It will increase 3 units for each unit to the right.

The point of intersection of the graphed lines is (2, 2).

The solution is (x, y) = (2, 2).

7 0

2 years ago