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4 years ago

Find the value of x in each case. Give reasons to justify your solutions! NEEDED ASAP

Mathematics

1 answer:

Mashcka [7]4 years ago

4 0

Answer:

( x = 17° )

Step-by-step explanation:

Consider the following steps;

$m< ZYW = 2x - Given,\\m< YXW = ( 3x - 5 ) - Given,\\m< XYW = 90 - Given,\\\\m< YXW + m< XYW + m< YWX = 180 - Sum of Angles in Triangle,\\( 3x - 5 ) + 90 + m< YWX = 180,\\3x - 5 + 90 + m< YWX = 180,\\3x - 5 + m< YWX = 90,\\m< YWX = 90 - 3x - 5,\\m< YWX = 85 - 3x,\\\\m< ZYW = m< YWX - Alternate Interior Angles,\\2x = 85 - 3x,\\5x = 85,\\Conclusion ; ( x = 17 degrees )$

Solution ; ( x = 17° )

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4 1/2 ounces per 3/4 miles _____ ounces per 1 mile

adelina 88 [10]

Answer: 6

Step-by-step explanation:

To get 3/4 miles to 1 mile we must take the reciprocal of the number and multiply it.

$\frac{3}{4}$ × $\frac{4}{3} =1$

We need to multiply the 4 1/2 ounces by 4/3 too

$4\frac{1}{2}$ × $\frac{4}{3} =6$

4 1/2 ounces per 3/4 miles

6 ounces per 1 mile

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3 years ago

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Which expression is equivalent to 5(y + 3)? a.15y + 5 b.5y + 15 c.15y d.8y

kobusy [5.1K]

Multiply the 5 over the answer is b

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3 years ago

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How to divide 6,875 by .025

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Answer= 285000
How to= the number you are dividing by has a decimal move the decimal point all the way to the right counting the number of places you’ve moved it to. Then move the decimal point in the number you’re dividing the same number of places to the right

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3 years ago

Factorize 2a^2+7a-15

loris [4]

$\huge\bf \pink {\underline{Solution :-}}$

$2 {a}^{2} + 7a - 15$

$= 2 {a}^{2} + 10a - 3a - 15$

$= 2a(a + 5) - 3(a + 5)$

$= (a + 5)(2a - 3)$

$\sf \red{Hence, Answer \: is \: (a + 5)(2a - 3).}$

$\\$

$\bf \purple{ \underline{ Important \: Formulas \: for \: Factorization :-}}$

$• \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}$

$• \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

$• \: {a}^{2} - 2 ab + {b}^{2} = {(a - b)}^{2}$

$• \: {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} = {(a + b)}^{3}$

$• \: {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} = {(a - b)}^{3}$

$• \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$• \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$• \: (a + b)( {a}^{2} -ab + {b}^{2} ) = {a}^{3} + {b}^{3}$

$• \: (a - b)( {a}^{2} + ab + {b}^{2} ) = {a}^{3} - {b}^{3}$

$• \: {( \frac{a + b}{2} )}^{2} - ( {\frac{a - b}{2} } )^{2} = ab$

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3 years ago

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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev

Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that $\mu = 201.9, \sigma = 2.1$ .

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7$

This probability is 1 subtracted by the pvalue of Z when $X = 204.1$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{204.1 - 201.9}{0.7}$

$Z = 3.14$

$Z = 3.14$ has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

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3 years ago