Evaluate f(4) =6(4)+5<br> X=4<br><br><br><br><br> Need answers ASAP

A bank loaned out 12,000 part of it at a rate of 8% per year and the rest at 18% per year. If the interest recurved in one year

spin [16.1K]

Answer:

$11,600

Step-by-step explanation:

amount loaned at 8% and ($12,000 - x) = amount loaned at 18%

Write the equation: (change the percents to decimals: 8% = 0.08 and 18% = 0.18)

(0.08)x + (0.18)($12,000 - x) = $1,000

This expresses the total amount of earned interest in terms of the amounts of the loan, x and ($12,000 - x}.

0.08x + $2,160 - 0.18x = $1,000 Simplify and solve for x.

-0.1x = -$1,160 Divide both sides by -0.1

x = $11.600 The amount loaned at 8% interest.

5 0

3 years ago

Please help me!!!! No idea how to do this one.

Volgvan

Class A - IQ1 - 76 - IQ3 - 81 - IQR - 5
Class B - IQ1 - 75 - IQ3 - 82 - IQR - 7
Class B is more variated than Class A.

3 0

3 years ago

A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the each toss (which are independent), compute the

andreyandreev [35.5K]

Answer:

(a) $\frac{1}{18}$ (d) $\frac{1}{9}$

(b) $\frac{1}{2}$ (e) $\frac{1}{3}$

(c) $\frac{4}{9}$ (f) $\frac{1}{12}$

Step-by-step explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability of getting a sum of 11 is:

$P(Sum 11) =\frac{2}{36} \\=\frac{1}{18}$

(b)

The sample space of getting an even sum is:

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6) (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6) (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

The probability of getting an even sum is:

$P(Even Sum)=\frac{18}{36}\\ =\frac{1}{2}$

(c)

The sample space of getting an odd sum more than 3 is:

{(1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3) and (6, 5)}

The probability of getting an odd sum more than 3 is:

<em>P</em> (Odd Sum more than 3) =

$=\frac{16}{36}\\=\frac{4}{9}$

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

$P(Even E1) =\frac{4}{36} \\=\frac{1}{9}$

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6} F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

$P(E>2, F$

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

$P(E=4, E+F = Odd)=\frac{3}{36} \\=\frac{1}{12}$

6 0

3 years ago

Use the quadratic formula to solve the equation.

Sergeeva-Olga [200]

ANSWER

$x = \frac{ 5 - \sqrt{ 5} }{4} \: or \: \: x = \frac{ 5 + \sqrt{ 5} }{4}$

EXPLANATION

The given quadratic equation is

$4 {x}^{2} - 10x + 5 = 0$

We compare this to

$a {x}^{2} + bx + c = 0$

to get a=4, b=-10, and c=5.

The quadratic formula is given by

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

We substitute these values into the formula to get:

$x = \frac{ - - 10 \pm \sqrt{ {( - 10)}^{2} - 4(4)(5)} }{2(4)}$

This implies that

$x = \frac{ 10 \pm \sqrt{ 100 - 80} }{8}$

$x = \frac{ 10 \pm \sqrt{ 20} }{8}$

$x = \frac{ 10 \pm2 \sqrt{ 5} }{8}$

$x = \frac{ 5 \pm \sqrt{ 5} }{4}$

The solutions are:

$x = \frac{ 5 - \sqrt{ 5} }{4} \: or \: \: x = \frac{ 5 + \sqrt{ 5} }{4}$

7 0

3 years ago

Read 2 more answers

Please help help help

3241004551 [841]

The value of the mean is 3 for normal curve C and the value of the mean is 4 for the normal curve D, and for both data standard deviation is the same.

<h3>What is a normal distribution?</h3>

It's the probability curve of a continuous distribution that's most likely symmetric around the mean. On the Z curve, at Z=0, the chance is 50-50. A bell-shaped curve is another name for it.

We have a normal distribution curve shown in the picture.

For the normal curve C:

The value of mean = 3 (draw a vertical line in the middle of the curve)

For the normal curve D:

The value of mean = 4 (draw a vertical line in the middle of the curve)

For both data standard deviation is the same.

Thus, the value of the mean is 3 for normal curve C and the value of the mean is 4 for the normal curve D, and for both data standard deviation is the same.

Learn more about the normal distribution here:

brainly.com/question/12421652

#SPJ1

5 0

2 years ago

Evaluate f(4) =6(4)+5 X=4 Need answers ASAP