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3 years ago

Need help don’t know what to do I’m confused

Mathematics

1 answer:

german3 years ago

4 0

It is the first one I can’t explain it but yea it’s the first choice.

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jim and lisa had a total of $300. After jim boughy a jacket with 2/3 if her money,she had $90 less than lisa. find the cost of t

Yuri [45]

Answer:

$84

Firstly, what we must do, is establish a system of equations:

$\frac{2}{3}J = L -90$

$J + L = 300$

Due the circumstances, it is decided we will use the method of substitution:

$\frac{2}{3}J = L - 90$

$\frac{2}{3}J + 90 = L$

↓

$J + \frac{2}{3}J + 90 = 300$

Now, we solve the 2-variable equation above:

$J + \frac{2}{3}J + 90 = 300$

$J + \frac{2}{3}J = 300 - 90$

$J + \frac{2}{3}J = 210$

$J(3) + (\frac{2}{3}J)(3) = 210(3)$

$3J + 2J = 630$

$5J = 630$

$J = \frac{630}{5}$

$\bf~J = 126$

Now, we must see how much money each one of them had & the price of the jacket:

Now we know that Jim had $126. In other words, $126 is $\frac{3}{3}$ of his money (total money).

So, if Jim has $126, which corresponds to $\frac{3}{3}$ of his money, $\frac{2}{3}$ of his money is:

$\frac{126}{3} = 42$

$126 * \frac{2}{3} = 42 + 42$

$\bf~126 * \frac{2}{3} = 84$

Hope it helped,

BiologiaMagister

3 0

3 years ago

Calc BC Problem. No random answers plz

stepladder [879]

Answer:

Part A)

$f(1)=2, \; f^{-1}(1)=0, \; f^\prime(1)=1.4, \; (f^{-1})^\prime(1)=\frac{10}{7}$

Part B)

$y=\frac{5}{14}x+\frac{4}{7}$

Step-by-step explanation:

Please refer to the table of values.

Part A)

A. 1)

We want to find f(1).

According to the table, when x=1, f(x)=2.

Hence, f(1)=2.

A. 2)

We want to find f⁻¹(1).

Notice that when x=0, f(x)=1.

So, f(0)=1.

Then by definition of inverses, f⁻¹(1)=0.

A. 3)

We want to find f’(1).

According to the table, when x=1, f’(x)=1.4.

Hence, f’(1)=1.4.

A. 4)

We will need to do some calculus.

Let g(x) equal to f⁻¹(x). Then by the definition of inverses:

$f(g(x))=x$

Take the derivative of both sides with respect to x. On the left, this will require the chain rule. Therefore:

$f^\prime(g(x))\cdot g^\prime(x)=1$

Solve for g’(x):

$g^\prime(x)=\frac{1}{f^\prime(g(x))}$

Substituting back f⁻¹(x) for g(x) yields:

$(f^{-1})^\prime(x)=\frac{1}{f^\prime({f^{-1}(x)})}$

Therefore:

$(f^{-1})^\prime(1)=\frac{1}{f^\prime({f^{-1}(1)})}$

We already determined previously that f⁻¹(1) is 0. Therefore:

$(f^{-1})^\prime(1)=\frac{1}{f^\prime(0)}$

According to the table, f’(0) is 0.7. So:

$(f^{-1})^\prime(1)=\frac{1}{0.7}=\frac{10}{7}$

Hence, (f⁻¹)’(1)=10/7.

Part B)

We want to find the equation of the tangent line of y=f⁻¹(x) at x=4.

First, let’s determine the points. Since f(2)=4, this means that f⁻¹(4)=2.

Hence, our point is (4, 2).

We will now need to find our slope. This will be the derivative at x=4. Therefore:

$(f^{-1})^\prime(4)=\frac{1}{f^\prime({f^{-1}(4)})}$

We know that f⁻¹(4)=2. So:

$(f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}$

Evaluate:

$(f^{-1})^\prime(4)=\frac{1}{f^\prime(2)}=\frac{1}{2.8}=\frac{10}{28}=\frac{5}{14}$

Now, we can use the point slope form. Our point is (4, 2) and our slope at that point is 5/14.

So:

$y-2=\frac{5}{14}(x-4)$

Solve for y:

$y-2=\frac{5}{14}x-\frac{20}{14}$

Adding 2 to both sides yields:

$y=\frac{5}{14}x-\frac{20}{14}+\frac{28}{14}$

Hence, our equation is:

$y=\frac{5}{14}x+\frac{4}{7}$

7 0

3 years ago

PLEASE HELP! 50 POINTS! Consider the following set of sample data: (34, 32, 34, 32, 40, 37, 31, 31, 29, 27). Which of the follow

Georgia [21]

Answer:

(15.23,41.016)

Step-by-step explanation:

WE must determine the mean of the data set: Which is the sum of the set divided by the number in the set.
$= (21 + 24 + 25 + 32 + 35 + 31 + 29 + 28)/8 = 225/8 = 28.125$
We must also determine the standard deviation: Which is the square root of the variance and the variance is the sum of squares of the sample number minus the mean divided by the number if the set data:
$= ((21 - 28.125)^{2} + (24 - 28.125)^{2} +(25 - 28.125)^{2} + (32 - 28.125)^{2} + (35 - 28.125)^{2} + (31 - 28.125)^{2} + (29 - 28.125)^{2} (28 - 28.125)^{2}$
$= 148.877/8 = 18.6$
The 95% confidence interval is defined as: The mean ± 1.96*standard deviation divided by the sqaure root of the number of data in the set:
$= 28.125 + (1.96 *18.6)/(\sqrt{8} )$
$= 41.016$
$= 28.125 - (1.96 * 18.6)/(\sqrt{8}) = 15.23$
The confidence interval for this data set is (15.23,41.016)