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2 years ago

Pls help and show workings

Mathematics

1 answer:

NemiM [27]2 years ago

3 0

The answer is C. The slope formula, y2-y1 over x2-x1, which can be remembered as rise over run, can be used to solve. The y2 is 7 as it is the second y and then 1, the first y, is subtracted from it. 7-1=6. After this, you find the second x which is 3, and subtract -6 from it, meaning that you add it as you are subtracting a negative and they cancel out. You now have 3-(-6), or 3+6 on the bottom and 7-1 on the top. 7-1=6 and 3+6=9. The answer is the 6/9 and can be simplified down to 2/3 because of their shared greatest common factor, 3. Hope this helps.

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2 years ago

Aimi is making Valentine’s Day cards for everyone in her class. She plans to use a whole sheet of paper for each of her 6 closes

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3 years ago

A carpenter has less than 120 min. to spend painting furniture each day. Today, he has spent 30.5 min. painting a desk. Now he w

FinnZ [79.3K]

Answer:

Hence, a carpenter can paint a maximum of 7 chairs.

Hence, option A is correct.

Step-by-step explanation:

A carpenter has less than 120 min. to spend painting furniture each day.

Today, he has spent 30.5 min. painting a desk.

Now he will paint x chairs, each of which takes 12.5 min.

i.e. he will take a total of 12.5x minutes in painting 'x' chairs.

Also total time spent on painting furniture each day will be equal to time spent on painting a desk and all of the chairs.

Hence,

30.5+12.5x<120 (since the total time he spends on painting is less than 120 mins.)

⇒ 12.5 x<120-30.5

⇒ 12.5 x<89.5

⇒ x<7.16 ( on dividing both side of the inequality with 12.5)

Hence, a carpenter can paint a maximum of 7 chairs.

6 0

3 years ago

Read 2 more answers

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musickatia [10]

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3 years ago

Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 2,250,000 and a mean

Svet_ta [14]

Answer: 0.1357

Step-by-step explanation:

Given : Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of $\sigma^2=2,250,000$ and a mean life span of $\mu=13,000$ hours.

Here , $\sigma=\sqrt{2250000}=1500$

Let x represents the life span of a monitor.

Then , the probability that the life span of the monitor will be more than 14,650 hours will be :-

$P(x>14650)=P(\dfrac{x-\mu}{\sigma}>\dfrac{14650-13000}{1500})\\\\=P(z>1.1)=1-P(z\leq1.1)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\=1-0.8643339=0.1356661\approx0.1357$

Hence, the probability that the life span of the monitor will be more than 14,650 hours = 0.1357

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3 years ago