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2 years ago

Which of following sets is equivalent to (a,b,c,d,e)

Mathematics

1 answer:

borishaifa [10]2 years ago

8 0

Answer:

I don't see what your trying too explain to me

I only see the question itself

Step-by-step:

sorry I can't help you

You might be interested in

Find the length of arc XW. Round your answer to the nearest hundredth.

Luda [366]

Answer:

15.08 m

Step-by-step explanation:

The angle XZW :

XZW + 108° = 180°

XZW = 180° - 108°

XZW = 72°

Length of an arc = θ / 360 * πd

d = diameter = XV = 24

Length of arc XW = (72/360) * π * 24

Length of arc XW = 15.0796

Length of XW = 15.08 m

4 0

3 years ago

16. Solve z2 – 2z – 35 = 0 using the zero product property.

kramer

<span>z2 – 2z – 35 = 0
2z - 2z = 35
0 = 35

No solution</span>

8 0

3 years ago

If 25% of a number is 12, what is 75% of that number?

snow_tiger [21]

Answer:

Step-by-step explanation:

$\frac{0.25}{0.75} =\frac{12}{x} \\$

$0.75 *12=9$

$\frac{9}{0.25} =36$

8 0

3 years ago

The diameter of a circle is 22 miles. What is the circle's area? Use 3.14 for pi.. Single choice.

Sonbull [250]

Answer:

379.94 mi²

Step-by-step explanation:

The area of a circle can be figured out with this formula: πr² (3.14 is π here, as it says in the directions).

We know the diameter is 22 miles. Divide that by 2 to get a radius of 11 miles.

The equation now becomes

3.14(11²)

11 squared is 121.

3.14(121)

I recommend splitting the 121 into 100, 20, and 1.

(3.14 * 100) + (3.14 * 20) + (3.14 * 1)

Multiply 3.14 by 100, 20, and 1.

314 + 62.8 + 3.14

Add 314 and 62.8.

376.8 + 3.14

Add 376.8 and 3.14

379.94 mi² is the area of a circle with a 22-mile diameter.

6 0

3 years ago

If x has a continuous uniform distribution on the interval (0,5), what is the 25th percentile for x?

Natasha_Volkova [10]

Given that $X\sim\mathrm{Unif}(0,5)$ , it has PDF

$f_X(x)=\begin{cases}\dfrac15&\text{for }0\le x\l5\\\\0&\text{otherwise}\end{cases}$

and thus CDF

$F_X(x)=\mathbb P(X\le x)=\begin{cases}0&\text{for }x$

The 25th percentile is the value $X=k$ such that $\mathbb P(X\le k)=0.25$ . We have

$\mathbb P(X\le k)=\dfrac k5=0.25\implies k=1.25$

4 0

3 years ago