Mr. Gonzalez bought 5 pounds of grapes for $15.How much did he spend per pound?

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago

PLEASE HELP asap it’s for a quiz if you know at least one answer please tell me

VARVARA [1.3K]

15
about 94.25
about 706.86

Lmk if you want to know how I got the answers

3 0

2 years ago

The National Association of College and University Business Officers collects data on college endowments. In 2015, its report in

Vikentia [17]

Answer:

.

Step-by-step explanation:

4 0

2 years ago

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s

jasenka [17]

Answer:

The minimum score of those who received C's is 67.39.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 73, \sigma = 11$

If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?

This is X when Z has a pvalue of 1-0.695 = 0.305. So it is X when Z = -0.51.

$Z = \frac{X - \mu}{\sigma}$

$-0.51 = \frac{X - 73}{11}$

$X - 73 = -0.51*11$

$X = 67.39$

The minimum score of those who received C's is 67.39.

7 0

3 years ago

Housing costs are what percentage of monthly net income? Use the budget sheet to calculate. Round your answer to the nearest who

schepotkina [342]

Answer:

See below

Step-by-step explanation:

The formula for calculating percent is

% = (cost of item/total cost) × 100 %

% = (cost of item/1875) × 100 %

% = (cost of item/18.75) %

You can use the same formula to calculate the percent for each item in the budget. For example,

Housing: % = 420/18.75 = 22 %

You can do this for each budget item and get the table below.

$\begin{array}{lrr}\text{Expense} & \text{Cost} & \%\\\text{Housing} & 430 & 22\\\text{Food} & 300 & 16\\\text{Utilities} & 125 & 7\\\text{Transportation} & 320 & 17\\\text{Savings} & 500 & 27\\\text{Other} & 200 & 11\\\text{TOTAL} & 1875 & 100\\\end{array}$

4 0

2 years ago