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3 years ago

Serena added 10 grams of sugar into 20 grams of water. She made _ grams of _% sugar syrup

2 answers:

7 0

Answer: She (Serena) made 30 grams of 33.33% sugar syrup. Solution: 10 grams sugar + 20 grams water As 10 grams of sugar added to 20 grams of water, total mixture became 30 grams. Percentage of sugar in total mixture (sugar syrup) = 10 grams sugar in 30 grams syrup. % by mass = [(mass of solute) / (mass of solute + mass of solvent)] * 100 This gives = [(10) / (10+20)] * 100 = 33.33%

aivan3 [116]3 years ago

7 0

She made 30 grams of 50% sugar syrup.

10 grams is 50% of 20 grams. Its 30 grams of syrup but ratio water:sugar is 2:1 so it's 50% syrup.

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Virty [35]

Answer: The mean is 3780

Step-by-step explanation:

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3 years ago

Identify the percent of change as an increase or a decrease. 1/4 to 1/2

rusak2 [61]

Increase.

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3 years ago

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EleoNora [17]

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3 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

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3 years ago

Is 48, 45, 41, 38, 34,... an arthmetic sequence

Evgesh-ka [11]

Answer:

Step-by-step explanation:

The change in the sequence is not consistent therefore it cannot be an arithmetic sequence.

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2 years ago

Serena added 10 grams of sugar into 20 grams of water. She made ___ grams of ___% sugar syrup

Serena added 10 grams of sugar into 20 grams of water. She made _ grams of _% sugar syrup