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3 years ago

Chuck can afford a $490-per-month car payment, and he's interested in either

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

3 0

Answer:

Chuck can afford both the convertible and sports car APEX

Step-by-step explanation:

Schach [20]3 years ago

3 0

Answer:

We can say that Chuck can afford both the cars.

Step-by-step explanation:

The EMI formula is :

$\frac{p*r*(1+r)^{n} }{(1+r)^{n}-1 }$

p = 28700

r = 6/12/100=0.005

n = 6*12=72

Putting the values in the above formula, we get:

$\frac{28700*0.005*(1.005)^{72} }{(1.005)^{72}-1 }$

Monthly payment is = $475.67

p = 29200

r = 6/12/100=0.005

n = 6*12=72

Putting the values in the above formula, we get:

$\frac{29200*0.005*(1.005)^{72} }{(1.005)^{72}-1 }$

Monthly payment is = $483.96

We can see that in both the cases, the monthly payments or EMI's are less than $490.

Therefore, we can say that Chuck can afford both the cars.

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yulyashka [42]

Answer:

Step-by-step explanation:

12(3/2)=36/2=18

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3 years ago

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Find the value of the expression below when x = 27, y = - 4 , and z = 64 . 2 * root(3, x) - 5sqrt(x) + y ^ 2 answer

Anna71 [15]

Answer:

Step-by-step explanation:

x = 27, y = - 4, z = 64

<u>Expression:</u>

2∛x - 5√z + y² =
2∛27 - 5√64 + (-4)² =
2(3) - 5(8) + 16 =
6 - 40 + 16 =
-18

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3 years ago

in a canoe race, a team paddles downstream 480 m in 60 s. The same team makes the trip upstream in 80 s. Find the team's rate in

Vera_Pavlovna [14]

480=60(v+s)

480=60v+60s

8=v+s

s=8-v...

480=80(v-s) using s from above we get:

480=80(v-(8-v))

6=v-8+v

6=2v-8

2v=14

v=7, and since s=8-v, s=1

So the rate of the team is 7m/s while the rate of the stream is 1m/s

7 0

3 years ago

Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

$\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out the sin²(x) from the denominator:

$\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (perfect square trinomial):

$\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

3 0

3 years ago

John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart

Rina8888 [55]

<h2>Answer:</h2>

$Probability=\frac{1}{24}$

<h2>Step-by-step explanation:</h2>

As the question states,

John's brother has Galactosemia which states that his parents were both the carriers.

Therefore, the chances for the John to have the disease is = 2/3

Now,

Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.

Now, one of those children married with a person.

So,

Probability for the child to have disease will be = 1/2

Now, again the child's child (Martha) probability for having the disease is = 1/2.

Therefore,

<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

$Probability=\frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\\Probability=\frac{1}{24}$

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)

<em><u>Hence, the probability for the first child to have Galactosemia is $\frac{1}{24}$ </u></em>

3 0

3 years ago