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3 years ago

The original price of a motorcycle was reduced by $275. The new discounted price is $1329.

1 answer:

7 0

Is there more to the question?

If not, then you are being asked to find the original price. If the price was reduced by $275, then add them back to the discount price.

$1329+$275= $1,604 is the original price

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A segment in the complex plane has a midpoint at –1 + i. If the segment has an endpoint at –5 – 7i, what is the other endpoint?

morpeh [17]

Answer:

The other end point is: s+ti = 3+9i

Step-by-step explanation:

Mid-Point(M) in the complex plane states that the midpoint of the line segment joining two complex numbers a+bi and s+ti is the average of the numbers at the endpoints.

It is given by: $M = \frac{a+s}{2} +(\frac{b+t}{2})i$

Given: The midpoint = -1 + i and the segment has an endpoint at -5 - 7i

Find the other endpoints.

Let a + bi = -5 -7i and let other endpoint s + ti (i represents imaginary )

Here, a = -5 and b = -7 to find s and t.

then;

$-1+i = \frac{-5+s}{2} + ( \frac{-7+t}{2})i$ [Apply Mid-point formula]

On comparing both sides

we get;

$-1 = \frac{-5+s}{2}$ and $1 = \frac{-7+t}{2}$

To solve for s:

$-1 = \frac{-5+s}{2}$

-2 = -5+s

Add 5 to both side we have;

-2+5 = -5+s+5

Simplify:

3 = s or

s =3

Now, to solve for t;

$1 = \frac{-7+t}{2}$

2 =-7+t

Add 7 to both sides we get;

2+7 = -7+t+7

Simplify:

9 = t

t =9

Therefore, the other end point (s+ti) is, 3+9i

4 0

4 years ago

Read 2 more answers

Of a squirrel's hidden nuts, for every 5 that get found, there are 3 that do not get found. A squirrel hid 40 nuts all together.

koban [17]

Answer:

15 nuts do not get found.

Step-by-step explanation:

Given that Of a squirrel's hidden nuts, for every 5 that get found, there are 3 that do not get found.

Total number of nuts squirrel had hidden = 40

Proportion of nuts not found to total = 3/(3+5) = 3/8

Hence out of 40, nuts not found =3/8(40) = 15 nuts.

15 nuts would not be found and 25 nuts would be found if squirrel had hidden in total 40 nuts.

This is because the proportion of found:unfound = 5:3

Hence 25:!5 =5:3 satisfies this

15 nuts are not found.

5 0

3 years ago

Read 2 more answers

A mixture of compounds X and Y in a 0.100-cm cell had an absorbance of 0.215 at 272 nm and 0.191 at 327 nm. Find [X] and [Y] in

Oksi-84 [34.3K]

Answer: The concentration of X is $7.13582\times 10^{-6}$ and the concentration of Y is $2.53159\times 10^{-5}$ .

Step-by-step explanation:

Since we have given that

At 272 nm, absorbance = 0.215

At 327 nm, absorbance = 0.191

As we have given that

Compound X Compound Y

272 16400 3870

327 3990 6420

So, our equations becomes

$16400C_1+3870c_2=0.215\\\\3990C_1+6420C_2=0.191$

By solving these two equations, we get that

$C_1=7.13582\times 10^{-6}\\\\C_2=2.53159\times 10^{-5}$

Hence, the concentration of X is $7.13582\times 10^{-6}$ and the concentration of Y is $2.53159\times 10^{-5}$ .

5 0

4 years ago

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

nordsb [41]

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$

5 0

4 years ago

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago