Ask question

Ask question

All categories

4 years ago

8

Mr. Johansen is cutting construction paper for an art project. He has 24 students, and six pieces of construction paper. How muc

h paper will each student receive?

1 answer:

wariber [46]4 years ago

5 0

Answer: One- fourth piece of construction paper is received by each student.

Step-by-step explanation:

Given: Mr. Johansen is cutting construction paper for an art project.

The number of students in the art project = 24

The number of pieces of construction paper = 6 pieces

When Mr. Johansen divides 6 pieces of construction paper in 24 students, the quantity of construction paper received by each student = $\frac{6}{24}=\frac{1}{4}\ \tex{piece}$

One- fourth piece of construction paper is received by each student.

You might be interested in

Find the margin of error for a 90% confidence interval when the standard deviation is LaTeX: \sigma= 50????=50 and LaTeX: n = 25

Murrr4er [49]

Answer:

The margin of error for a 90% confidence interval is 16.4

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 25

Standard deviation = 50

$z_{critical}\text{ at}~\alpha_{0.10} = \pm 1.64$

Margin of error =

$z_{critical}\times \dfrac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$1.64\times \dfrac{50}{\sqrt{25}} = 16.4$

Thus, the margin of error for a 90% confidence interval is 16.4

8 0

4 years ago

Classify the following triangle. check all that apply.

podryga [215]

Answer:

Scalene and acute

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Brainiest goes to the first correct answer thanks

quester [9]

Answer:

20

Step-by-step explanation:

we can see that there are already answers for 2 which is six 6+6=12 now we can see that a triangle look the same as the six over the 8mm so that must be six so 6+6+6=18 now there is still s little section missing over the 8mm (the half triangle half square) is missing a section we can see that the perimeter is 8 and the area is 6 and 6+2=8 so that little area is 2 so 6+6+6+2=20

6 0

4 years ago

The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

$N\sim Pois(0.2)$

$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0

4 years ago

8. An Isosceles triangle HIJ has sides HI and HJ as 19 meters each. If angle I and angle J are 67 degrees, what is the distance

PtichkaEL [24]

Answer:

$IJ=14.8$

Step-by-step explanation:

In a triangle the sum of all angle measures is always 180 degrees

$A+B+C=180$

$2(67)+C=180$

$C=46$ degrees

By splitting the isosceles triangle down the middle we can calculate the base

So ∠H = 23°, ∠K = 90°, ∠J = 67°, HJ = 19

With this we can solve for half of the base. the trig function will need to use is sin, since the side opposite of ∠H will be part of the base. $sin=\frac{opposite}{hypotenuse}$

now we need to rearrange the formula for what we have.

$hypotenuse(sin)=opposite$

$19sin(23)=7.423$

This value is only half of the base, so we need to double it

$IJ=2(7.423)=14.8$

8 0

3 years ago