1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
FrozenT [24]
3 years ago
14

For the parallelogram, if m<2=5x-28 and m<4=3x-10, find m<3​

Mathematics
1 answer:
Yuliya22 [10]3 years ago
3 0
5x-28=3x-10
X=9
5*9-28=40
360-40-40=280
280/2=140=m<3
You might be interested in
James earns 20 allowance every month and 30 for each a on his report card
Svetach [21]

Answer:

Part A - the independant variable is the monthly allowance, and the dependent amount is the one that he has to get As for in order to receive it.

Part B - y = 30x + 20

Step-by-step explanation:

7 0
3 years ago
Help please homework due tomorrow and I don't know how to do this
Furkat [3]
Number 1 is 0.2 and 0.4
number 2 is 0.53 and 0.56
number 3 is on the 0.3
number 4 is 1/3
number 5 is a little bit before the 0.7 mark
number 6 is 62/100
4 0
3 years ago
Read 2 more answers
Simplify the expression. (5x2 - x - 7) + (2x2 + 3x + 4)
kifflom [539]

Answer:11+2x

Step-by-step explanation:

(10-7-x)+(4+4+3x)

3-x+8+3x

3+8+3x-x

11+2x

3 0
4 years ago
A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective
Inessa05 [86]

Answer:

(a) P(X \leq 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

            r = number of success

           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

Now, the z score probability distribution is given by;

                Z = \frac{X-\mu}{\sigma} ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X \leq 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

5 0
3 years ago
Word Problem
Sladkaya [172]

Step-by-step explanation:

1 l = 22.5 km

125 l = 22.5 × 125 = 2812.5 km

5 0
3 years ago
Other questions:
  • Mr.Church had 3 muffins he sold 2 but then baked 6 more and sold 4 more how many muffin did he not sell.
    5·2 answers
  • Im distributive property and combining like term and I have to simplify this expression 10(7r)
    15·1 answer
  • 5. Solve v = w - 20 for w.​
    6·2 answers
  • FIRST TO ANSWER, would be marked as brainliest:)
    14·2 answers
  • Is 5 a solution to -3x=15
    6·2 answers
  • Please understand the explanation
    9·1 answer
  • Please help I don't get it
    8·1 answer
  • What do the graphs of sine and cosine have in common with the swinging you see?
    5·1 answer
  • Plzzzzzzzzzzzzzzzzzzzzzzzz
    7·1 answer
  • Y= 7x - 19<br> y= 3x - 7<br><br> Solve for x and then use X to solve for y. (find x and y)
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!