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2 years ago

Can someone pls help me

Mathematics

2 answers:

Novosadov [1.4K]2 years ago

8 0

9(2n + 1)

Distribute 9 to the terms in the brackets.

18n + 9

Add = 0 to solve for x

18n + 9 = 0

Move 9 to the other side by subtracting

18n = -9

Divide by 18 on both sides

n = -9/18

Simplify

n = -1/2

In decimal form the answer can also be -0.5 depending on your question.

Hope this helped!

anygoal [31]2 years ago

6 0

Answer:

Sorry that I cant help.

But have a good rest of your dayFollow if you liked this.

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PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

saw5 [17]

Answer: A

Step-by-step explanation: A

6 0

3 years ago

Discuss his principles of inheritance by answering the following questions. Two plants are being crossed, each heterozygous for

disa [49]

Answer:

Q1 - What is the possibility that the offspring will have white flowers? Express your answer as a ratio.

A: White flowers only appear if the offspring is homozygous for b because this phenotype is recessive, so they have to present bb phenotype. If both parents are heterozygous (Bb), each one gives one gene at a time per individual after crossing, so 1/4 would be BB, 2/4 would be Bb and 1/4 would be bb.

Q2 - If one parent is also heterozygous for smooth peas, does this mean that all of the offspring with blue flowers will also have smooth peas? Explain your answer without using a Punnett square.

A: No, it doesn't mean that all blue flowers will present smooth peas because those genes are expressed in different loci, so they segregate independently. There might be some blue flowers with smooth peas, but that is not mandatory for all of them.

Q3 - Suppose both parents are also heterozygous for smooth peas (S), with wrinkled peas (s) as the recessive trait. Use a Punnett square to predict the possibility that the offspring will be homozygous for both flower color and pea texture. Make sure to include a Punnett square and express your prediction as a ratio.

A: Parental genotypes are BbSs x BbSs. Given that the genes segregate independently, we will have the Punnett diagram as follows:

BS / Bs / bS / bs

BS BBSS / BBSs / BbSS / BbSs

Bs BBSs / BBss / BbSs / Bbss

bS BbSS / BbSs / bbSS / bbSs

bs BbSs / Bbss / bbSs / bbss

Homozygous genotypes are the ones that both genes are the same when expressed, either BB,bb,SS or ss. The probability the offspring is homozygous for both flower color and pea texture is 4/16, or 1/4.

5 0

3 years ago

In a purely monopolistic market with a demand curve P = -Q / 10 + 2000, to maximize profit the firm provides the application at

Rudiy27

Answer: hello your question lacks some data hence I will be making an assumption to help resolve the problem within the scope of the question

answer:

≈ 95 units ( output level )

Step-by-step explanation:

Given data :

P = 2000 - Q/10

TC = 2Q^2 + 10Q + 200 ( assumed value )

<u>The output level where a purely monopolistic market will maximize profit</u>

P = 2000 - Q/10 ------ ( 1 )

PQ = 2000Q - Q^2 / 10 ( aka TR )

MR = d (TR ) / dQ = 2000 - 2Q/10 = 2000 - Q/5

TC = 2Q^2 + 10Q + 200 ---- ( 2 )

MC = d (TC) / dQ = 4Q + 10

equating MR = MC

2000 - Q/5 = 4Q + 10

2000 - 10 = 4Q + Q/5

1990 = 20Q + Q

∴ Q = 1990 / 21 = 94.76 ≈ 95 units ( output level )

7 0

2 years ago

The holding period of property acquired by gift may begin on:

maw [93]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago