All categories

2 years ago

I'd appreciate it if anyone helped! :)

Mathematics

2 answers:

ra1l [238]2 years ago

6 0

At 0 weeks she had 75 L water remaining in the barrel (0, 75)

Since she used 5L each week, then at 15 weeks she had 0 L remaining (15, 0).

Answer: Top Right graph

Illusion [34]2 years ago

3 0

Answer:

Graph given on the top right side.

Step-by-step explanation:

Nina used 5 L of water from her rain barrel to water her plants every week.

She did this until her rain barrel was empty.

So barrel will be empty in the time = $\frac{75}{5}=15$ weeks.

Nina started watering when rain barrel contained 75L of water.

We draw a graph with water remaining on y-axis and weeks to get the barrel empty on x-axis.

Line representing this process will start from a point (0, 75) on y-axis and will end at x-axis at the point (15, 0).

Therefore, graph given on the top right side of the options is the correct answer.

You might be interested in

If y varies directly as x, and y=-4 when x=8, find y when x=3

enot [183]

I believe it would be negative 9 because your subtracting by twelve again. I hope that helped

3 0

3 years ago

10. Which of the following situations represents a proportional relationship?....

tia_tia [17]

Answer:

G A pool fills at a rate of 90 gallons per hour

Step-by-step explanation:

i dont know

7 0

2 years ago

Here are 180 candies in a candy shipment. There are Gobs, Blobs, and Toffees. The bage are made in ratio of 4:2:3 of Gobs, to Bl

Maru [420]

Answer:

The answer to the cost for each toffee is given below:

Step-by-step explanation:

The computation of the cost for each toffee is shown below:

Given that

There are a total 180 candies hat includes three types in a ratio of 4:2:3

Gobs = 4 ÷ 9 × 180 = 80

Blobs = 2 ÷ 9 × 180 = 40

Toffees = 3 ÷ 9 × 180 = 60

Now the total cost of all the candies shipment is $430

So,

The equation would be

80 × $1.50 + 40 × $2.50 + 60 × per cost = $430

120 + 150 + 60 × per cost = $430

60 × per cost = $430 - $120 - $150

60 × per cost = $160

Per cost is $2.67

So for toffees it would be 160

4 0

2 years ago

What is the equation of the line?

Elina [12.6K]

Answer:

A or $\frac{1}{2}x +2$

Step-by-step explanation:

Remember y=mx+b | m = slope & b = y intercept

Think back of rise over run. Where the rise is 1 and 2 is run. Now Also, lets not forget the y intercept which you can easily obtain by looking where the line intercepts the y axis. Which in this case it would be 2.

3 0

3 years ago

Read 2 more answers

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

$Z_{1-\alpha /2}= Z_{0.95}= 1.648$

0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

0.48±1.648* $\sqrt{\frac{0.48*0.52}{1000} }$

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: $\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015$

Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0

3 years ago