A cookie recipe for 5 dozen cookies calls for 4 cups of flour. How many is flour is needed to make 90 cookies

Anybody know the answers to this frl it’s due in 15 mins

hammer [34]

Answer:

4. j(11) = 8

I think it is linear

5. g(-3) = 18

No, it's a parabola since x is to the second power

6. f(-7) = 5

No, I think it's an absolute value function

7. h(0) = 13

I think it is linear, just the slope is 0

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What are some benefits of representing data sets using frequency distributions?

nekit [7.7K]

Answer:

no answer

Step-by-step explanation:

no answer

to be checked in options

3 0

1 year ago

One out of 3 gym members say they spent 5 hours per week at the gym. If there are 150 people working out at the gym, how many wi

lana [24]

Answer: 50

Step-by-step explanation:

From the question, we are informed that 1/3 of gym members say they spent 5 hours per week at the gym.

Assuming there are 150 people working out at the gym, the number of people that'll most likely spend 5 hours at the gym this week will be calculated by multiplying 1/3 by 150. This will be:

= 1/3 × 150

= 50 people

8 0

3 years ago

The length of a rectangle is four times its width. If the area of the rectangle is 324m^2, find its perimeter.

Anuta_ua [19.1K]

Answer:

$perimeter=90$

Step-by-step explanation:

We know that the length is four times the width, so:

$l=4w$

We also know the area, which is 324 m². The formula for area:

$A=l*w$

Insert the known values:

$324=(4w)*w$

Solve for w. Simplify by removing parentheses:

$324=4w*w\\324=4w^2$

Divide 4 from both sides to isolate the variable:

$\frac{324}{4}=\frac{4w^2}{4} \\\\81=w^2$

Find the square root of both sides:

$\sqrt{81} =\sqrt{w^2} \\\\w=9$

The width is 9 m.

We know the width. Now find the length by using the area formula and inserting known values:

$324=l*9$

Solve for l. Divide both sides by 9:

$\frac{324}{9}=\frac{l*9}{9}\\\\ l=36$

The length of the rectangle is 36. (You can check: 4 times 9 is 36)

Now find the perimeter:

$P=2l+2w$

Insert values:

$P=2(36)+2(9)\\\\P=72+18\\\\P=90$

The perimeter is 90 m.

5 0

3 years ago

A data mining routine has been applied to a transaction dataset and has classified 88 records as fraudulent (30 correctly so) an

Firlakuza [10]

Answer:

The classification matrix is attached below

Part a

The classification error rate for the records those are truly fraudulent is 65.91%.

Part b

The classification error rate for records that are truly non-fraudulent is 96.64%

Step-by-step explanation:

The classification matrix is obtained as shown below:

The transaction dataset has 30 fraudulent correctly classified records out of 88 records, that is, 30 records are correctly predicted given that an instance is negative.

Also, there would be 88 - 30 = 58 non-fraudulent incorrectly classified records, that is, 58 records are incorrectly predicted given that an instance is positive.

The transaction dataset has 920 non-fraudulent correctly classified records out of 952 records, that is, 920 records are correctly predicted given that an instance is positive.

Also, there would be 952 - 920 = 32 fraudulent incorrectly classified records, that is, 32 records incorrectly predicted given that an instance is negative.

That is,

Predicted value

Active value Fraudulent Non-fraudulent

Fraudlent 30 58

non-fraudulent 32 920

The classification matrix is obtained by using the information related to the transaction data, which is classified into fraudulent records and non-fraudulent records.

The error rate is obtained as shown below:

The error rate is obtained by taking the ratio of $\left( {b + c} \right)(b+c)$ and the total number of records.

The classification matrix is, shown above

The total number of records is, $30 + 58 + 32 + 920 = 1,040$

The error rate is,

$\begin{array}{c}\\{\rm{Error}}\,{\rm{rate}} = \frac{{b + c}}{{{\rm{Total}}}}\\\\ = \frac{{58 + 32}}{{1,040}}\\\\ = \frac{{90}}{{1,040}}\\\\ = 0.0865\\\end{array}$

The percentage is $0.0865 \times 100 = 8.65$

(a)

The classification error rate for the records those are truly fraudulent is obtained by taking the rate ratio of b and $\left( {a + b} \right)(a+b)$ .

The classification error rate for the records those are truly fraudulent is obtained as shown below:

The classification matrix is, shown above and in the attachment

The error rate for truly fraudulent is,

$\begin{array}{c}\\FP = \frac{b}{{a + b}}\\\\ = \frac{{58}}{{30 + 58}}\\\\ = \frac{{58}}{{88}}\\\\ = 0.6591\\\end{array}$

The percentage is, $0.6591 \times 100 = 65.91$

(b)

The classification error rate for records that are truly non-fraudulent is obtained by taking the ratio of d and $\left( {c + d} \right)(c+d)$ .

The classification error rate for records that are truly non-fraudulent is obtained as shown below:

The classification matrix is, shown in the attachment

The error rate for truly non-fraudulent is,

$\begin{array}{c}\\TP = \frac{d}{{c + d}}\\\\ = \frac{{920}}{{32 + 920}}\\\\ = \frac{{920}}{{952}}\\\\ = 0.9664\\\end{array}$

The percentage is, $0.9664 \times 100 = 96.64$

6 0

3 years ago