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2 years ago

NO LINKS!! Part 2Similarity Flow Chart Proofs

Mathematics

1 answer:

STatiana [176]2 years ago

4 0

Answer:

See below

Step-by-step explanation:

According to diagram we have

QR ≅ QT
QS ≅ QS (common side)
QSR ≅ QST (both right angles)

Considering above we can state

QSR ≅ QST by HL (hypotenuse-leg)

Two angles are congruent but the order of angles is not same

Triangles are not similar.

Two angles and a side are congruent but the order of angles is not same.

Triangles are not similar.

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Find the amount in an account where $500 is invested at 2.5% compounded continuously for period of 10 years

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500 *1.025^10 ≈ 640.04

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3 years ago

I have four bills that need to be paid this week. The electric bill is 182.17 the water bill is 92.55 the cell phone bill is 225

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The answer would be 200 + 200 +200+100 = 700

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3 years ago

Read 2 more answers

amid [387]

The correlation based on the computation is 0.97.

<h3>How to compute the value?</h3>

It should be noted that the question is that we should simply calculate the correlation which is r.

The correlation will be:

= Σ(x y) / (Σ(x²) × Σ(y²)

= 9627/✓(1118 × 87982)

= 9627 ✓98363876

= 9627/9917.86

= 0.97

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brainly.com/question/4658834

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1 year ago

Formulas can be manipulated to solve for missing information.

shepuryov [24]

The formulas answer/ variable?

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3 years ago

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of <em>X</em> as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

4 0

3 years ago

NO LINKS!! Part 2Similarity Flow Chart Proofs​

NO LINKS!! Part 2Similarity Flow Chart Proofs