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1 year ago

Sam and Edna have 56 marbles together Edna has 6 times more marbles than Sam. How many marbles does sam have ?

Mathematics

2 answers:

Elis [28]1 year ago

7 0

The answer is 8 marbles. 6*8=48. 48(ednas marbles) + 8(sam's marbles)=56.

nataly862011 [7]1 year ago

4 0

Answer: Sam 25 marbles.

Step-by-step explanation: Given that Sam and Edna have 56 marbles together and Edna has 6 times more marbles than Sam.

We are to find the number of marbles that Sam has.

Let x and y represents the number of marbles that Sam and Edna has respectively.

Then, according to the given information, we have

$x+y=56\\\\\Rightarrow y=56-x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y=x+6\\\\\Rightarrow 56-x=x+6\\\\\Rightarrow 2x=50\\\\\Rightarrow x=\dfrac{50}{2}\\\\\Rightarrow x=25.$

Thus, Sam has 25 marbles.

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Tanzania [10]

Answer:

Approximately, 159 men weighs more than 165 pounds and 159 men weighs less than 135 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 pounds

Standard Deviation, σ = 15

We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P( men weighing more than 165 pounds)

P(x > 165)

$P( x > 165) = P( z > \displaystyle\frac{165 - 150}{15}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 165) = 1 - 0.8413 = 0.1587 = 15.87\%$

Approximately, 159 men weighs more than 165 pounds.

P(men weighing less than 135 pounds)

P(x < 135)

$P( x < 135) = P( z < \displaystyle\frac{135 - 150}{15}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 135) = 0.1587 = 15.87\%$

Approximately, 159 men weighs less than 135 pounds.

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2 years ago

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Leokris [45]

Answer:

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Explanation:

If Jason were to be paid per catch given that he makes a total of 110 catches for the 2017 season in his new contract, he would make a total of :

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1 year ago

Caleb draws 14 dogs on each of 4 posters.He draws 18 cats on each of 6 other posters. If he draws 5 more dogs on each poster wit

MatroZZZ [7]

Caleb draws 14 dogs on each of 4 posters.
Then,
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Then
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Again Caleb draws 5 more dogs on each poster with dogs.
Then,
The number of Dogs drwn in the second instance = (5 * 4) dogs
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So, the total number of dogs drawn by Caleb = ( 56 + 20) dogs
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2 years ago

Consider a data set with at least three data values. Suppose the highest value is increased by 10 and the lowest is decreased by

bulgar [2K]

Answer:

a. Mean doesn't change.

b. Median doesn't change.

c. Mode can change.

Step-by-step explanation:

Let us assume the data set with 10 observations

{2,6,4,3,2,6,4,9,4,7}.

Arranging data set in ascending order

{2,2,3,4,4,4,6,6,7,9}

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median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

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median=(4+4)/2=8/2=4

Mode

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Increasing highest value by 10 and decreasing lowest value by 10

{-8,2,3,4,4,4,6,6,7,19}

mean=-8+2+3+4+4+4+6+6+7+19/10=4.7

Mean doesn't change

median

n/2=10/2=5 is an integer so,

median= average of n/2 and n/2 +1

median= (5th value+6th value)/2

median=(4+4)/2=8/2=4

Median doesn't change

Most occurring value is still 4. But mode can change if the value the highest value becomes most concurring value.

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