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3 years ago

Nathan rolls a number cube and records the result of each roll in the table.

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

True statements:

A. The relative frequency of rolling 4 is $\dfrac{2}{9}.$

E. The experimental probability of rolling 1 is less than the experimental probability of rolling 6.

Step-by-step explanation:

Nathan made 11 + 16 + 14 + 20 + 12 + 17 = 90 rolls altogether. Consider all options:

A. The relative frequency of rolling 4 is

$\dfrac{20}{90}=\dfrac{2}{9}.$

This option is true.

B. The experimental probability of rolling 3 is $\dfrac{14}{90}.$ The theoretical probability of rolling 3 is $\dfrac{1}{6}=\dfrac{15}{90}.$ therefore, the experimental probability of rolling 3 is less than the theoretical probability. This option is false.

C. The experimental probability of rolling 2 is $\dfrac{16}{90}$ and the theoretical probability of rolling 2 is $\dfrac{1}{6}=\dfrac{15}{90}.$ The experimental probability of rolling 2 is smaller than the theoretical probability of rolling 2. Option C is false.

D. The relative frequency of rolling 5 is $\dfrac{12}{90}=\dfrac{2}{15}\neq \dfrac{2}{13}.$ This option is false.

E. The experimental probability of rolling 1 is $\dfrac{11}{90}$ and the experimental probability of rolling 6 is $\dfrac{17}{90}.$ Since $\dfrac{11}{90}$ then the experimental probability of rolling 1 is less than the experimental probability of rolling 6. This option is true.

F. The theoretical probability of rolling 1 is the same as the theoretical probability of rolling 6. This option is false.

dimulka [17.4K]3 years ago

6 0

Answer:

A and E

Step-by-step explanation:

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 29 ft/s. Its height

Ilia_Sergeevich [38]

Answer:

$\overline{v}_{@\Delta t=0.01s}=-15.22ft/s, \overline{v}_{@\Delta t=0.005s}=-15.11ft/s, \overline{v}_{@\Delta t=0.002s}=-15.044ft/s, \overline{v}_{@\Delta t=0.001s}=-15.022ft/s$

Step-by-step explanation:

Now, in order to solve this problem, we need to use the average velocity formula:

$\overline{v}=\frac{y_{f}-y_{0}}{t_{f}-t_{0}}$

From this point on, you have two possibilities, either you find each individual $y_{f}, y_{0}, t_{f}, t_{0}$ and input them into the formula, or you find a formula you can use to directly input the change of times. I'll take the second approach.

We know that:

$t_{f}-t_{0}=\Delta t$

and we also know that:

$t_{f}=t_{0}+\Delta t$

in order to find the final position, we can substitute this final time into the function, so we get:

$y_{f}=29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}$

so we can rewrite our formula as:

$\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-y_{0}}{\Delta t}$

$y_{0}$ will always be the same, so we can start by calculating that, we take the provided function ans evaluate it for t=1s, so we get:

$y_{0}=29t-22t^{2}$

$y_{0}=29(1)-22(1)^{2}$

$y_{0}=7ft$

we can substitute it into our average velocity equation:

$\overline{v}=\frac{29(t_{0}+\Delta t)-22(t_{0}+\Delta t)^{2}-7}{\Delta t}$

and we also know that the initil time will always be 1, so we can substitute it as well.

$\overline{v}=\frac{29(1+\Delta t)-22(1+\Delta t)^{2}-7}{\Delta t}$

so we can now simplify our formula by expanding the numerator:

$\overline{v}=\frac{29+29\Delta t-22(1+2\Delta t+\Delta t^{2})-7}{\Delta t}$

$\overline{v}=\frac{29+29\Delta t-22-44\Delta t-22\Delta t^{2}-7}{\Delta t}$

we can now simplify this to:

$\overline{v}=\frac{-15\Delta t-22\Delta t^{2}}{\Delta t}$

Now we can factor Δt to get:

$\overline{v}=\frac{\Delta t(-15-22\Delta t)}{\Delta t}$

and simplify

$\overline{v}=-15-22\Delta t$

Which is the equation that will represent the average speed of the ball. So now we can substitute each period into our equation so we get:

$\overline{v}_{@\Delta t=0.01s}=-15-22(0.01)=-15.22ft/s$

$\overline{v}_{@\Delta t=0.005s}=-15-22(0.005)=-15.11ft/s$

$\overline{v}_{@\Delta t=0.002s}=-15-22(0.002)=-15.044ft/s$

$\overline{v}_{@\Delta t=0.001s}=-15-22(0.001)=-15.022ft/s$

5 0

3 years ago

Simplify: (8^2)^3<br> please help:)

alexira [117]

Answer:

262144

Step-by-step explanation:

its exponents

6 0

3 years ago

Read 2 more answers

What is the distance between (4, 3) and (9, 15) on the coordinate plane?

nalin [4]

Answer:

169

Step-by-step explanation:

3 0

3 years ago

Choose the correct mathematical translation for the given statement.

Lunna [17]

X less than 70. not really sure what your asking but I'm pretty sure this is correct

6 0

3 years ago

Find the equation of the median from b in ABC whose vertices are (1,5), B(5,3) and C(-3, -2)

Soloha48 [4]

Answer:

y = x + 6

x = 1

y = ¼(x - 5) + 3

Step-by-step explanation:

Vetices are;

A(1,5), B(5,3) and C(-3, -2)

Thus;

Median of AB is; D = (1 + 5)/2, (5 + 3)/2

D = (3, 4)

Median of BC is; E = (5 + (-3))/2, (3 + (-2))/2

E = (1, 0.5)

Median of AC is; F ; (-3 + 1)/2, (-2 + 5)/2

F = (-1, 1.5)

Thus, the median lines will be;

CD, AE & BF.

Thus;

Equation of CD is;

(y - (-3))/(x - (-2)) = (-2 - 4))/(-3 - 3)

(y + 4)/(x + 2) = -6/-6

y - 4 = 1(x + 2)

y = 4 + x + 2

y = x + 6

Equation of AE;

(y - 5)/(x - 1) = (0.5 - 5)/(1 - 1)

(y - 5)/(x - 1) = -4.5/0

Cross multiply to get;

0(y - 5) = -4.5(x - 1)

-4.5x = -4.5

x = 1

Equation of BF;

(y - 3)/(x - 5) = (1.5 - 3)/(-1 - 5)

(y - 3)/(x - 5) = -1.5/-6

(y - 3)/(x - 5) = 1/4

y - 3 = ¼(x - 5)

y = ¼(x - 5) + 3

7 0

3 years ago