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2 years ago

Which option please I need help ASAP reply ASAP

Mathematics

2 answers:

IrinaK [193]2 years ago

8 0

Answer:

The answer is probably c

Step-by-step explanation:

its deducted by the x axis

Dafna1 [17]2 years ago

5 0

Answer:

Step-by-step explanation:

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I need HELP ASAP! It’s due today

Vikki [24]

Answer:

28 cubs

Step-by-step explanation:

Since one and a half bear produces one and a half cub, then 6 bears will produce 6 cubs in one and a half days.

We need to find 7 days, though but we have the rate for six bears for 1.5 days which 6 bears one a half day.

What is the rate for a day?

6/1.5 =

6 bears produce 4 cubs a day.

Now just do 4 x 7

4 x 7 = 28

Therefore, 6 bears produce 28 cubs in 7 days.

8 0

2 years ago

Read 2 more answers

PLS HELP 50 POINTS

natali 33 [55]

Answer:

f(x+1)=(5/6)f(x)on: Five-sixths Superscript f (x)

EXPLANATION:

The defining characteristic of a geometric sequence is that each term is a constant multiple of the previous term, called the common ratio denoted r in the rules for geometric equations.

625/750=750/900=etc=r=5/6

4 0

2 years ago

Read 2 more answers

Find sin θ if θ is in Quadrant III and tan θ = . 0.958

Leokris [45]

Use the following identities:
$sec^2 = 1 + tan^2 \\ \\ sec = \frac{1}{cos} \\ \\ sin^2 = 1 - cos^2$
Also because the angle is in quadrant 3, sin must be negative.
Therefore
$sin = - \sqrt{1 - \frac{1}{1 + tan^2}}$
Subbing in tan = 0.958
$sin \theta = -0.69178$

7 0

2 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

The difference of 5 times y and 3

saul85 [17]

The way to find the difference between 5y - 3 you have to first do the inverse operation of multiplication which is division.

5y - 3
__ __
5 5

y - 0.6

3 0

2 years ago