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3 years ago

Edward has 30 sweets.

Mathematics

2 answers:

posledela3 years ago

4 0

Answer:

Step-by-step explanation:

Sum the parts of the ratio, 2 + 3 = 5 parts

Divide the number of sweets by 5 to find the value of one part of the ratio

30 ÷ 5 = 6 sweets ← value of 1 part of the ratio, thus

2 × 6 = 12 ← number of red sweets

lianna [129]3 years ago

3 0

Answer:

20 red sweets

Step-by-step explanation:

do two thirds of 30 and the part that cane out is the red sweets

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Your parents took your family out to dinner. Your parents wanted to give the waiter a 15% tip. If the total amount of the dinner

Novosadov [1.4K]

Answer:

$48.30

Step-by-step explanation:

If the dinner was $42.00 and they wanted to give a 15% tip, then to find that start by multiplying 42 by 0.15

42.00 x 0.15 = 6.3

now add $6.30 to $42.00

42 + 6.3 = 48.3

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3 years ago

Read 2 more answers

The clock was exactly on time at 7am at 1pm the clock was 228 seconds late at that rate how slow was the clock half an hour

yan [13]

Answer:

19 seconds.

Step-by-step explanation:

Given that the clock was exactly on time at 7am, and at 1pm the clock was 228 seconds late, to determine, at that rate, how slow was the clock half an hour, the following calculation must be performed:

1 PM = 13:00

13 - 7 = 6

228/6 = 38

38/2 = 19

Thus, every half hour the clock is delayed 19 seconds.

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3 years ago

The diameter of a circle is 14 in. Find its area in terms of

Luda [366]

Answer:

Given - Diameter of circle = 14 in

To find - Area of circle

Solution -

Diameter= 2 × Radius

14 = 2 × R

14/2 = R

7 = R

Area of circle= π r^²

22/7 × 7 × 7

22 × 7

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3 years ago

Calculator

polet [3.4K]

Answer:

Step-by-step explanation:

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3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

4 years ago