How do you get that m

Gwent pour 256 milliliters of water in to Ravi Henry pours 189 milliters of water in to Ravis .Ravis now contains 800 Milliters

Contact [7]

Answer:

Ravi had 355ml water to begin with.

Step-by-step explanation:

Given that,

amount of water that Gwent pour in Ravi = 256 ml

amount of water that Henry pour in Ravi = 189 ml

amount of water after each have poured = 800 ml

suppose amount of water ravi had before both poured water in it = x

Equation

256 + 189 + x = 800

445 + x = 800

x = 355ml

Initially ravi had 335ml of water.

6 0

3 years ago

An Italian restaurant offers small, medium and large calzones. The choices of filings are cheese, sausage, spinach, or vegetable

larisa86 [58]

Answer:

A) 12 toppings.

Step-by-step explanation:

That would be the answer if there are no combinations of toppings.

You would simply take the number of sizes (3) and the number of toppings (4) and multiply them together to get 12 combinations.

8 0

3 years ago

At a school fair 70% of the people are under 16 years old. One third of the people remaining are teachers. If there are 21 teach

Tamiku [17]

In the fairest school 70% are below 16 years old 1/3 are teachers which is equals to = 21 Let’s start solving: => 1/3 of 100% => 100 / 3 = 33.33% thus 33.33% = 21 => 21 x 3 = 63, is the total number of people in the school. Let’s try solving the number of people below 16 years old Have you notice that you are asking for a 70% of students but there are already 33.33% of teacher. Thus your given problem is not right already. => 100% - 33.33% = 66.67% that’s the only remaining percentage and not 70% => 63 * .6667 = 42.0021 Thus, there are around 42 people who are 16 years old younger.

3 0

3 years ago

Steel Factory Workers Ages

igor_vitrenko [27]

Solution: We are given the population mean $=42$

Now, in order to find which shift's mean is closest to population mean, we will find the mean of each shift.

The mean of shift 1 is:

$Mean=\frac{18+25+56+42+29+38+54+47+35+30}{10}=\frac{374}{10}=37.4$

The mean of shift 2 is:

$Mean=\frac{23+19+50+49+67+34+30+59+40+33
}{10}=\frac{404}{10}=40.4$

The mean of shift 3 is:

$Mean=\frac{19+22+24+40+45+29+33+29+39+59 }{10}=\frac{339}{10}=33.9$

The mean of shift 4 is:

$Mean=\frac{21+23+25+40+35+19+70+40+22+23 }{10}=\frac{318}{10}=31.8$

We clearly see the mean of shift 2 is close to the population mean. Hence the option B) Shift 2 is correct.

7 0

3 years ago

Read 2 more answers

If a cat with short brown hair (llBb) is crossed with a cat with long black hair (Llbb), what is the likelihood that there will

Advocard [28]

Bb = brown hair phenotype

bb = black hair phenotype

we see that B = brown hair is the dominant gene while b = black hair is the recessive gene

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ll = short hair, this is a lowercase L

Ll = long hair

In this case, l = short hair is the recessive gene while L = long hair is the dominant gene.

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Check out the diagram below to see the dihybrid crosses for each of the hair color and hair length genotypes. From there, I show all of the genotypes possible for any offspring. Those genotypes are

LlBb = long brown hair (comes up 4 times)
Llbb = long black hair (comes up 4 times)
llBb = short brown hair (comes up 4 times)
llbb = short black hair (comes up 4 times)

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If we are talking about 1 offspring, then it is not possible for him/her to have both "long black hair AND short brown hair" as the instructions state. So let's consider two offspring instead.

The probability of having long black hair is 4/16 = 1/4 as there are 4 cases of Llbb out of 16 cases total. Similarly, the probability of getting llBb is 4/16 = 1/4 as well, for pretty much the same reasons.

The probability of having both events occur is (1/4)*(1/4) = 1/16

The multiplication of probabilities is valid because the two events are independent.

==============================================

<h3>Final Answer: A) 1/16</h3>

5 0

2 years ago