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2 years ago

Algebra. Find the variable in the equation 803 divided by 73=m

Mathematics

1 answer:

weeeeeb [17]2 years ago

8 0

<u>Answer:</u>

The value of variable "m" in the equation 803 divided by 73 = m is equal to 11.

<u>Solution:</u>

Need to find the variable in the given equation 803 divided by 73 = m

That means need to solve below expression to determine value of variable m.

$\frac{803}{73}=m$

So let’s solve left hand side to get the value of m.

For sake of simplicity let’s do factorization of numerator and denominator and remove the common terms in numerator and denominator.

$\begin{array}{l}{803=73 \times 11} \\\\ {73=73 \times 1} \\\\ {=>\frac{803}{73}=\frac{73 \times 11}{73 \times 1}=11}\end{array}$

Hence value of variable m is equal to 11.

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Evelyn has $524.96 in her checking account. She must maintain a $500 balance to avoid a fee. She wrote a check for $32.50 today.

barxatty [35]

Answer:

524.96 − 32.50 + x ≥ 500

Amount need to deposit = $7.54

Step-by-step explanation:

Given:

Amount in checking account = $524.96

Maintain amount = $500

Amount of check = $32.50

Find:

Amount need to deposit

Computation:

Assume, amount need to deposit = x

So,

For avoiding fee

524.96 − 32.50 + x ≥ 500

x = 7.54

Amount need to deposit = $7.54

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2 years ago

Clarissa's income puts her in the bottom tax bracket (10%) last year. During the same year, she earned $250 in dividends and $75

Art [367]

Answer:

75 in coupons.

250 in dividends.

profit of 600 - 425 = 175 from her stock investment.

her total income is 250 + 75 + 175 = 500.

if all of this is taxed at 10%, then her tax will be 500 * .1 = 50.

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2 years ago

A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation

Bas_tet [7]

Answer:

Option b. None is the correct option.

The Answer is 63%

Step-by-step explanation:

To solve for this question, we would be using the z score formula

The formula for calculating a z-score is given as:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

We have boxes X and Y. So we will be combining both boxes

Mean of X = 100 lb

Mean of Y = 5 lb

Total mean = 100 + 5 = 105lb

Standard deviation for X = 1 lb

Standard deviation for Y = 0.5 lb

Remember Variance = Standard deviation ²

Variance for X = 1lb² = 1

Variance for Y = 0.5² = 0.25

Total variance = 1 + 0.25 = 1.25

Total standard deviation = √Total variance

= √1.25

Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,

For 104lb

z = (x-μ)/σ,

z = 104 - 105 / √25

z = -0.89443

Using z score table ,

P( x = z)

P ( x = 104) = P( z = -0.89443) = 0.18555

For 1061b

z = (x-μ)/σ,

z = 106 - 105 / √25

z = 0.89443

Using z score table ,

P( x = z)

P ( x = 106) = P( z = 0.89443) = 0.81445

P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555

= 0.6289

Converting to percentage, we have :

0.6289 × 100 = 62.89%

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Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%

Since there is no 63% in the option, the correct answer is Option b. None.

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