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3 years ago

The table shows the relationship "Mason rode 15 miles per hour on his bicycle."

Mathematics

2 answers:

Tems11 [23]3 years ago

4 0

The statements which are correct are-

The number of miles increases as time increases.
The number of hours causes a change in the number of miles ridden.
The variable h is the independent variable.

Step-by-step explanation

The above table displays the relationship between the time taken in hours ("h") by Mayon to cover the distance (miles "m").

It means that the distance is dependent on time in hours. Hence, the variable "m" is a dependent variable since it depends on time "h" for which bicycle is being ridden.

Similarly, the variable "h" is an independent variable.

As seen from the table, the value of "m" continuously increases for the value of "h" and thus the number of hours causes a change in the distance ridden.

Lunna [17]3 years ago

4 0

Answer:

The number of miles increases as time increases.

The number of hours causes a change in the number of miles ridden.

The variable h is the independent variable.

The variable m is the dependent variable.

Step-by-step explanation:

Your welcome e d g e n u i t y people

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Answer:

We conclude that the average height of hobbit is taller than Yoda.

Step-by-step explanation:

We are given that in the Star Wars franchise, Yoda stands at only 66 centimetres tall.

From a sample of 7 hobbits, you ﬁnd their mean height $\bar X$ = 80 cm with standard deviation s = 10.8 cm.

Let $\mu$ = average height of hobbit.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 66 cm {means that the average height of hobbit is shorter than or equal to Yoda}

Alternate Hypothesis, $H_A$ : $\mu$ > 66 cm {means that the average height of hobbit is taller than Yoda}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean height = 80 cm

s = sample standard deviation = 10.8 cm

n = sample of hobbits = 7

So, test statistics = $\frac{80-66}{\frac{10.8}{\sqrt{7}}}$ ~ $t_6$

= 3.429

The value of t test statistics is 3.429.

Now, at 0.01 significance level the t table gives critical value of 3.143 for right-tailed test. Since our test statistics is more than the critical value of t as 3.429 > 3.143, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average height of hobbit is taller than Yoda.

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4 years ago

What happens to the value of the expression b-1 as b increases

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Answer:

It increases

Step-by-step explanation:

if b increases and the only other part of the equation is a constant it should increase as well. for example (the first value is b and the second is b-1) : (1,0) (2,1) (3,2) (4,3) (5,4)...

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3 years ago

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i have 17 coins. N of them are nickels and the rest are dimes. write an expression in two different ways for the amount of money

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Answer:

Step-by-step explanation:

(N)0.05 + (17-N)0.1 = M; M = amount of money I have.

Or 1.7-0.05N = M.

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3 years ago

Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein

ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

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Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

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k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

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