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Westkost [7]
3 years ago
15

Gianna is preparing for the upcoming tennis season. According to her schedule, she can complete at most 10 workouts this week. I

t takes Gianna 45 minutes to complete her drill-based workouts and 30 minutes to complete her conditioning workouts. She is required to spend more than 450 minutes on workouts every week. Which system of inequalities can be used to determine the number of drill-based workouts, x, and the number of conditioning workouts, y, that Gianna can complete this week?
Mathematics
1 answer:
Strike441 [17]3 years ago
4 0
The whole number of minutes that Gianna complete each week as a function of x and y:
45x+30y
Since the above number is greater than 450 minutes, we get the inequality:
45x+30y\geq450
Also, "she can complete at most 10 workouts this week" so we deduce the inequality:
\\x+y\leq10
Therefore, the system of inequalities is the following:
45x+30y\geq450
\\x+y\leq10
The above system has many solutions, for example :
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Consider the scatter plot.
Leno4ka [110]

<u>Answer:</u>

The correct answer option is: The y-intercept of the line of best fit shows that when time started, the distance was 5 feet.

<u>Step-by-step explanation:</u>

We are given a scatter plot with a best fit line as shown on the given graph.

The equation of the best fit line is given by:

y = 0.75x + 5

So with the help of the equation and by looking at the given graph, we can conclude about the representation of the y intercept that the the y-intercept of the line of best fit shows that when time started, the distance was 5 feet.

Since the distance shown on the y axis is already 5 when the time started at 0 minutes.

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3 years ago
2) Write an equation that passes through the<br> point (-2,5) and is parallel to y ==x+1.
Marta_Voda [28]

Answer:

y = x + 7

Step-by-step explanation:

5 = -2 + b

7 = b

y = x + 7

* Parallel lines have SIMILAR <em>RATE</em><em> </em><em>OF</em><em> </em><em>CHANGES</em><em> </em>[<em>SLOPES</em>], so 1 remains the way it is.

I am joyous to assist you anytime.

4 0
3 years ago
The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.30​, and the sample statistics
IrinaK [193]

Answer: 0.84

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let p be the population proportion of adults who smoked a cigarette in the past week.

As per given , we have

H_a: p

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The test statistic for proportion is given by :-

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

Substitute all the values , we get

z=\dfrac{0.31-0.30}{\sqrt{\dfrac{0.30(1-0.30)}{1491}}}\\\\=\dfrac{0.01}{\sqrt{0.000141}}\\\\=\dfrac{0.01}{0.011874342087}=0.84261497731\approx0.84

Hence, the value of the test statistic = 0.84

6 0
3 years ago
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