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2 years ago

Kathy has 3/4 of a yard of fabric. She needs 3/10 of a yard for each doll dress she makes. How many doll dresses can she make?

1 answer:

4 0

In order to calculate correctly, it is important that we are able to understand well the given values. We have
3/4 yard fabric = total amount of available fabric
3/10 yard= amount needed to make one doll
Therefore,
number of doll dresses = 3/4 / 3/10 = 2 doll dresses

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The table shows data gathered by an environmental agency about the decreasing depth of a lake during the first few months of a d

EastWind [94]

(0,346)(2,344.8)
slope = (344.8 - 346) / (2 - 0) = -1.2 / 2 = -0.6

y = mx + b
slope(m) = -0.6
use either of ur points (0,346)...x = 0 and y = 346
now we sub and find b, the y int
346 = -0.6(0) + b
346 = b

so ur equation is : y = -0.6x + 346

after 4 weeks....x = 4
y = -0.6(4) + 346
y = -2.4 + 346
y = 343.6 <=== after 4 weeks it will be 343.6

5 0

2 years ago

-6/5-2/5v+4/15+1/3v please help

PtichkaEL [24]

Answer:

-14/15-1/15v

Step-by-step explanation:

First you make all of the fractions have the common denominator 15.

And then you have -18/15-6/15v+4/15+5/15v

Then you combine like terms to get -14/15-1/15v

5 0

2 years ago

Sociologists want to test whether the number of homeless people in a particular urban area is increasing. In 2010, the average n

Kobotan [32]

Answer:

The test statistic value is 15.3.

Step-by-step explanation:

The hypothesis for this test is:

H₀: The average number of homeless people is not increasing, i.e. μ = 42.3.

Hₐ: The average number of homeless people is increasing, i.e. μ > 42.3.

Given:

$\bar x=45.3\\\sigma=6.2\\n=1000$

As the population standard deviation is provided use a single mean z-test for the hypothesis testing.

The test statistic is:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{45.3-42.3}{6.2/\sqrt{1000}}=15.3$

Thus, the test statistic value is 15.3.

7 0

2 years ago

Dahlia is trying to decide which bank she should use for a loan she wants to take out. In either case, the principal of the loan

vovangra [49]

Answer:

a. Loan Q’s finance charge will be $83.73 greater than Loan P’s.

Step-by-step explanation:

Loan P

Monthly payments for a 9-year 5.8% loan for $19,450 are $231.59, so the total repaid is ...

108 payments × $231.59 per payment = $25,011.72

Together with the service charge, the cost of the loan is ...

$25,011.72 + 925.00 - 19,450 = $6,486.72

Loan Q

Monthly payments for a 10-year 5.5% loan for $19,450 are $211.08, so the total repaid is ...

120 payments × $211.08 per payment = $25,329.60

Together with the service charge, the cost of the loan is ...

$25,329.60 +690.85 -19,450 = $6,570.45

Difference of charges

The cost of Loan Q exceeds that of Loan P by ...

$6,570.45 -6486.72 = $83.73 . . . . . matches choice A

_____

The payment values here were computed using a financial calculator. They can be found using the TVM Solver of a graphing calculator (TI-84), or by any spreadsheet. A number of web sites will also calculate loan payment values, or you can use the formula in your text to compute them yourself.

3 0

2 years ago

The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds prod

True [87]

Answer:

a) 57.35%

b) 99.99%

c) 68.27%

Step-by-step explanation:

When we have a random variable X that is normally distributed with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ , then

The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the left of a.

The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the right of b.

The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤ b) = P(a < X ≤ b)= P(a ≤ X < b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ between a and b.

In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml

(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?

As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).

You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.

If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula:

1- NORMDIST(62,63,5.4,1) in Excel

1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc

and we will get a value of 0.5735 or 57.35%

(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?

Now we want the area to the left of 90

NORMDIST(90,63,5.4,1) in Excel

NORMDIST(90;63;5.4;1) in OpenOffice Calc

You will get a value of 0.9999 or 99.99%

(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?

You can use either the rule that 68.27% of the data falls between $\large\bf \mu -\sigma$ and $\large\bf \mu +\sigma$ or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4

In Excel

NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1)

In OpenOffice Calc

NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1)

In any case we get a value of 0.6827 or 68.27%

3 0

2 years ago