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erik [133]
2 years ago
4

Ashley found 2 boxes of sugar in the kitchen.the green box says 1.26 kg and the red box says 1.026 kg.which box contains more su

gar
Mathematics
2 answers:
madam [21]2 years ago
4 0

Answer:the red box

Step-by-step explanation:

the red box since  1.26 is greater than 1.026, because the tenths place has the most power after a decimal. and the tenths place in the green box of sugar is 2, while the tenths place in the red box of sugar is 0 and 2 is greater than 0

hope this helps!

iogann1982 [59]2 years ago
3 0

Answer:

Green box contains more sugar.

Step-by-step explanation:

Consider the provided information.

Ashley found 2 boxes of sugar in the kitchen.

The green box says 1.26 kg and the red box says 1.026 kg.

we need to find which box contains more sugar.

To find this we will first check the digit at ones place.

Both the digit has 1 at the ones place. So now check the digit at tenths place.

1.26 has 2 at tenths place while 1.026 has 0 at tenths place.

As we know 2 is greater than 0, thus the number 1.26 is greater than 1.026.

Hence, green box contains more sugar.

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It represents a mosquito that flew very fast after feeding relative to all other mosquitoes.

Step-by-step explanation:

The point is unusual because the velocity of the mosquito is substantially greater than all other velocities and is probably an outlier.

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2 years ago
What is the multiplicative rate of change of the function? One-fifth Two-fifths 2 5
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A on edge

Step-by-step explanation:

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2 years ago
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Using a 16-sided number cube, what is the probability that you will roll a multiple of 4 or an odd prime number? The number 1 is
iogann1982 [59]
<h3>Answer: 0.563 (choice C)</h3>

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

Explanation:

E = event space

E = set of outcomes we want to happen

E = rolling a multiple of 4 or rolling an odd prime number

E = {4,8,12,16     3,5,7,11,13}

n(E) = number of items in set E

n(E) = 9, (four multiples of 16+five ways to roll odd prime number)

-----

S = sample space

S = set of all possible outcomes (whether we want them to happen or not)

S = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}

n(S) = number of items in set S

n(S) = 16

-----

P(E) = probability of event space occurring

P(E) = probability of rolling a multiple of 4 or rolling an odd prime number

P(E) = n(E)/n(S)

P(E) = 9/16

P(E) = 0.5625

P(E) = 0.563

3 0
2 years ago
Alexander invested $320 in an account paying an interest rate of 1.5% compounded annually. Assuming no deposits or withdrawals a
andrew11 [14]
I think it is
320 x (0,015^18) x 100
3 0
2 years ago
A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly locat
DiKsa [7]

Answer:

The probability of a  selection of 50 pages will contain no errors  is  0.368

The probability that the selection of the random pages will contain at least two errors is 0.2644

Step-by-step explanation:

From the information given:

Let q represent the no of typographical errors.

Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let \mu be the random variable that follows a Poisson distribution, then mean \mu = \dfrac{10}{500}= 0.02

and the mean that the random selection of 50 pages will contain no error is \lambda = 50 \times 0.02 =1

∴

Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}

Pr(q =0) = 0.368

The probability of a  selection of 50 pages will contain no errors  is  0.368

The probability that 50 randomly page contains at least 2 errors is computed as follows:

P(X ≥ 2) = 1 - P( X < 2)

P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )]    since it is less than 2

P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]

P(X \geq 2) = 1 - [0.3678 +0.3678]

P(X \geq 2) = 1 -0.7356

P(X ≥ 2) = 0.2644

The probability that the selection of the random pages will contain at least two errors is 0.2644

6 0
2 years ago
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