I need help with 1-2.b

Mashutka [201]

17. All
18. B because it can be symmetrical either way

3 0

3 years ago

Using a metric line as taught in your lesson, convert the following metric measurement. (Remember to use the same number of sign

weqwewe [10]

Answer:

Option B.

Step-by-step explanation:

We need to express 505 kL in L.

Here, kL means kiloliter and L means liter.

We know that

1 kL = 1000 L

Using this conversion we can easily convert 505 kL in L.

Multiply both sides by 505.

505 × 1 kL = 505 × 1000 L

505 kL = 505,000 L

The value of 505 kL can be expressed as 505,000 L.

Therefore, the correct option is B.

4 0

3 years ago

Read 2 more answers

A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi

SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

$= \frac{20!}{(20-5)! 5!)}$

$= \frac{20*19*18*17*16}{5*4*3*2*1}$

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

$= \frac{20}{320} = \frac{1}{19}$

P(A5) = ⁸C₅ / ²⁰C₅

$= \frac{56}{15540} = \frac{7}{1938}$

P(A4) = ⁷C₄ / ²⁰C₄

$= \frac{35}{4845} = \frac{7}{969}$

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0

3 years ago

Reduced ratio 27 to 9

Sidana [21]

3:1 is your answer :)

3 0

3 years ago

The scale from a square park to a drawing of the park is 5 miles to 1 miles. The actual park has an area of 1,600 m×2 what is th

Flauer [41]

The user corrected that the scale of a drawing of a park reads: 5 miles to 1 cm , and we know that the park measures 1,600 square meters (user insisted that this measure is given in square meters and not square miles).

So we have to convert the 1600 square meters into miles, knowing that 1 meter is the same as: 0.000621371 miles

then meters square will be equivalents to:

1 m^2 = (0.000621371 mi)^2

then 1600 m^2 = 0.00061776 mi^2

now, since 5 miles are represented by 1 cm, then 25 square miles will be represented by 1 square cm

and therefore 0.00061776 square miles will be the equivalent to:

0.00061776 / 25 cm^2 = 0.000024710 cm^2

So and incredibly small number of square cm.

I still believe that some of the information you gave me are not in meters but in miles. (For example, the park may not be in square meters but in squared miles). The park seems to have the size of a house according to the info.

3 0

1 year ago