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3 years ago

2 kg in the ratio 3:5

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

3 0

Answer:

3u--> 750g

5u-->1250g

Step-by-step explanation:

2kg = 2000g

3+5=8

8u --> 2000g

1u--> 250g

3u--> 750g

5u-->1250g

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A game contains 3 bags of green chips 2 bags of red chips and 5 bags og yellow chips. there ard 25 green chips in a bag and 30 r

nika2105 [10]

1 game has 3 bags of green chips ; 2 bags of red chips ; 5 bags of yellow chips
1 bag: 25 green chips
1 bag: 30 red chips

total number of chips:
green: 3 x 25 = 75
red: 2 x 30 = 60
yellow: 75 + 60 = 135 answer to part a.

75 + 60 + 135 = 270 total number of chips.

270 / 54 = 5 players. answer to part b.

Each of the 5 players will have 15 green chips, 12 red chips, and 27 yellow chips, for a total of 54 chips.

8 0

3 years ago

Could someone please show me how to figure this problem out?

elixir [45]

The baby blue whale weighs 30 pounds, since 120 divided by 40 is 30.

5 0

3 years ago

Rewrite the fractions 1/3 and 3/5 as fractions with a least common denominator

kifflom [539]

3, 6, 9, 12, 15

5, 10, 15

1/3 = 5/15

3/5 = 9/15

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3 0

3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

7 0

3 years ago

Mario invests 1,500 in a savings account that earns 2% interest a year. He also plans to set aside $50 cash a month. A:2100(1.02

Sedbober [7]

Answer:

<em>1500(1.02)^x + 600x</em> is how much he has in savings at the end of x years where it be in the bank or elsewhere

Step-by-step explanation:

x is in years

Let's just think about the investment of 1500 in an account earning 2% per year.

Before the years even start, you are at 1500 ( present value).

The next year (year 1), it would be 1500*.02+1500=(1500)(1.02).

The next year (year 2), it would be 1500(1.02)(.02)+1500(1.02)=1500(1.02)(1.02).

We keep multiplying factors of (1.02) each time.

So for year x, you would have saved 1500(1.02)^x.

Now we are saving 50 cash per month. Per year this would be 12(50) since there are 12 months in a year. 12(50)=600.

So the first year you would have 600.

The second year you would have 600(2) or 1200.

The third year you would have 600(3) or 1800.

Let's put this together:

1500(1.02)^x + 600x

4 0

3 years ago