All categories

3 years ago

On Wednesday, how many total kilograms of flour were prodľiced at plant #1 and

Mathematics

1 answer:

Svet_ta [14]3 years ago

8 0

Answer:

110,000 kg

Step-by-step explanation:

Just look at the kilos of flour produced on Wednesday from both plants and add them together :)

You might be interested in

ramone has 5 difficult questions left to answer on a multiple choice test. Each question has 3 choices. For the first 2 of these

olchik [2.2K]

1. a b c
2. a b c
3. a b c
4. a b c
5. a b c

then she eliminated 1 choice in 1 and 2, say as follows

1. b c
2. a b
3. a b c
4. a b c
5. a b c

Probability of answering correctly the first 2, and at least 2 or the remaining 3 is
P(answering 1,2 and exactly 2 of 3.4.or 5.)+P(answering 1,2 and also 3,4,5 )

P(answering 1,2 and exactly 2 of 3.4.or 5.)=
P(1,2,3,4 correct, 5 wrong)+P(1,2,3,5 correct, 4 wrong)+P(1,2,4,5 correct, 3 wrong)
also P(1,2,3,4 c, 5w)=P(1,2,3,5 c 4w)=P(1,2,4,5 c 3w )
so
P(answering 1,2 and exactly 2 of 3.4.or 5.)=3*P(1,2,3,4)=3*1/2*1/2*1/3*1/3*2/3=1/4*2/9=2/36=1/18

note: P(1 correct)=1/2
P(2 correct)=1/2
P(3 correct)=1/3
P(4 correct)=1/3
P(5 wrong) = 2/3

P(answering 1,2 and also 3,4,5 )=1/2*1/2*1/3*1/3*1/3=1/108

Ans: P= 1/18+1/108=(6+1)/108=7/108

5 0

3 years ago

Alicia answered 20 out of 25 problems correctly on a test. What percent did she get correct?

DedPeter [7]

80% of the problems correct.

3 0

3 years ago

Read 2 more answers

A hiker travels 12 km in 3 h.

FinnZ [79.3K]

This is a ratio.
12 km : 3 h

simplify to-
4km : 1 h

which means the hiker traveled 4 km per 1 hour.

6 0

3 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$