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2 years ago

Solve the Inequality for x x/2+3/2<5/2

Mathematics

1 answer:

charle [14.2K]2 years ago

7 0

Answer:

x<2

Step-by-step explanation:

I didnt feel like typing it out but the explanation is in the picture hope this helps :)

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I would like to ask about question d

Marrrta [24]

6 halves is 3.
B is at 3 1/2
3 1/2 minus 3 is 1/2
D would be plotted at 1/2.

7 0

3 years ago

A leaky faucet in kali's house drips water at a rate of 42 gallons per week. How many gallons does the faucet drip in three days

Ratling [72]

Amount of water that the faucet leaks in 1 week = 42 gallons

That means Amount of water that the faucet in Kali's house leaks in 7 days = 42 gallons So The amount of water that the faucet in Kali's house leaks in 1 day = (42/7) gallons = 6 gallons Then The amount of water that the faucet in Kali's house leaks in 3 days = 6 * 3 gallons = 18 gallons

So it can said from the above deduction that the amount of water leaked by the faucet in Kali's house is 18 gallons.

hope this help .-.

5 0

3 years ago

Factor the polynomial by its greatest common monomial factor.

Gwar [14]

Answer: 3b^4(b+5-6b^3)

Step-by-step explanation:

6 0

3 years ago

There are 135 people and three door prizes. how many ways can door prizes of $7,500, $750 and $75 be distributed?

igomit [66]

Answer: 2405970 ways

Step-by-step explanation:

When the order of selecting r individuals out of n individual , we use permutations.

The number of permutations of r things taking out of n things is given by :-

$^nP_r=\dfrac{n!}{(n-r)!}$

Given : There are 135 people and three door prizes .

i.e. n= 135 , r= 3

Then, the total number of ways = $^{135}P_3=\dfrac{135!}{(135-3)!}$

$=\dfrac{135\times134\times133\times132!}{132!}\\\\=135\times134\times133\\\\=2405970$

Hence, there are 2405970 ways to distribute door prizes of $7,500, $750 and $75 .

6 0

3 years ago

A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample

8090 [49]

Answer:

Step-by-step explanation:

Step(i):-

Given first random sample size n₁ = 500

Given Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

First sample proportion

$p^{-} _{1} = \frac{65}{500} = 0.13$

Given second sample size n₂ = 700

Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.

second sample proportion

$p^{-} _{2} = \frac{133}{700} = 0.19$

Level of significance = α = 0.05

critical value = 1.96

Step(ii):-

Null hypothesis : H₀: There is no significance difference between these proportions

Alternative Hypothesis :H₁: There is significance difference between these proportions

Test statistic

$Z = \frac{p_{1} ^{-}-p^{-} _{2} }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }$

where

$P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2} }{n_{1}+ n_{2} } = \frac{500 X 0.13+700 X0.19 }{500 + 700 } = 0.165$

Q = 1 - P = 1 - 0.165 = 0.835

$Z = \frac{0.13-0.19 }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }$

Z = -2.76

|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance

Null hypothesis is rejected at 0.05 level of significance

Alternative hypothesis is accepted at 0.05 level of significance

Conclusion:-

There is there is a difference between these proportions at α = 0.05

3 0

3 years ago