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3 years ago

1. m+ 10< 16

Mathematics

1 answer:

sleet_krkn [62]3 years ago

7 0

Answer:

$1.\ m$

Step-by-step explanation:

1. Subtract 10 from both sides. Then:

$m+ 10< 16\\ m+ 10-10< 16-10\\m$

2. Divide both sides by -2. Notice that the direction of the symbol of the inequality will change. Then:

$-2g\geq8\\\\\frac{-2g}{-2}\geq\frac{8}{-2}\\\\g\leq-4$

3. Add 22 to both sides. Then:

$y-22< 19\\ y-22+22$

4. Divide both sides by -7. Notice that the direction of the symbol of the inequality will change. Then:

$-7b>-28\\\\\frac{-7b}{-7}>\frac{-28}{-7}\\\\b$

5. Subtract 30 from both sides. Then:

$h+30$

6. Divide both sides by -5. Notice that the direction of the symbol of the inequality will change. Then:

$-5x$

7. Subtract 13 from both sides. Then:

$t+13>22\\ t+13-13>22-13\\t>9$

8. Add 12 to both sides. Then:

$w-12< 16\\ w-12+12$

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three sets of the sum of a number and four are added to the sum of seven times the same number and thirteen

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The algebraic expression for given statement is: 10x + 25 or 3(x + 4) + (7x + 13)

Solution:

Given the statement:

Three sets of a sum of a number and four are added to the sum of seven times the same number and thirteen

Let us first understand the given statement,

Let the number be "x"

" sum of a number and four" means x + 4

"Three sets of a sum of a number and four" translated to 3(x + 4)

"sum of seven times the same number and thirteen" means 7x + 13

Thus the algebraic expression for given statement is:

$3(x+4)+(7x + 13)$

Using distributive property in above expression

$a(b+c) = ab + bc$

Therefore,

$3(x+4)+(7x + 13) = 3x + 12 + 7x + 13$

Combine the like terms

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Thus the required expression for given statement is: 10x + 25 or 3(x + 4) + (7x + 13)

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3 years ago

A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies

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Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = weights of the fruits

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean weight = 733 grams

$\sigma$ = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

P(X > x) = 0.05 {where x is the required weight}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-733}{9}$ ) = 0.05

P(Z > $\frac{x-733}{9}$ ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

$\frac{x-733}{9}=1.645$

${x-733}}=1.645\times 9$

$x = 733 + 14.81$

x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

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