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2 years ago

Which additional congruence statement could you use to prove that RQS= PQS by hl

Mathematics

1 answer:

Nady [450]2 years ago

5 0

Answer:

The correct answer is last option SR ≅ SP

Step-by-step explanation:

Points to remember

IF two right angled triangles are congruent by HL congruence, then one leg and hypotenuse of both are congruent,

To find the correct option

From the figure we get,

SQ is the common side of two triangles.

SR is the hypotenuse of ΔRQS and

SP is the hypotenuse of ΔPQS

Therefore we need SR ≅ SP

The correct answer is last option SR ≅ SP

You might be interested in

Identify two numbers less than 20 with the most factors

Anastaziya [24]

Answer

18 and 12

Explanation

Factor is a number or algebraic expression that divides another number or expression evenly that is with no remainder.

Odd number have few factors so the can't be the answer.

WE are going to try the even numbers.

18 ⇒ 18, 9, 6, 3, 2, 1

16 ⇒ 16, 8, 4, 2, 1

12 ⇒ 12, 6, 4, 3, 2, 1

10 ⇒ 10, 5, 2, 1

Other number less than 10 has factors less than 5.

It can be seen from above that 18 and 12 have the most factors.

3 0

2 years ago

Read 2 more answers

A survey revealed that 31% of people are entertained by reading books, 39% are entertained by watching TV, and 30% are entertain

tester [92]

1% entertained by books
9% entertained by TV

3 0

2 years ago

Kathi and Robert Hawn had a pottery stand at the annual Skippack Craft Fair. They sold some of their pottery at the original pri

grigory [225]

Answer:27 pieces were sold at the original price.

63 pieces were sold at the new price

Step-by-step explanation:

Let x represent the number of pieces of pottery that was sold at the original price.

Let y represent the number of pieces of pottery that was sold at the new price.

They sold some of their pottery at the original price of $9.50 for each piece. This means that the amount that they got from selling x pieces of pottery at the original price would be 9.5x

They later decreased the price of each piece by $2. This means that the new price was 9.5 - 2 = $7.5

This means that the amount that they got from selling x pieces of pottery at the new price would be 7.5y

If they sold all 90 pieces and took in $729, then the equations are

x + y = 90

9.5x + 7.5y = 729 - - - - - - - - - -1

Substituting x = 90 - y into equation 1, it becomes

9.5(90 - y) + 7.5y = 729

855 - 9.5y + 7.5y = 729

- 9.5y + 7.5y = 729 - 855

- 2y = - 126

y = - 126/- 2 = 63

Substituting y = 63 into x = 90 - y, it becomes

x = 90 - 63 = 27

6 0

2 years ago

Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu

Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of $P(Z$ , you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0

2 years ago

A researcher took a random sample of 100 students from a large university. She computed a 95% confidence interval to estimate th

katen-ka-za [31]

Answer:

Part (A): The correct option is true.

Part (B): The null and alternative hypothesis should be:

$H_o: \mu =0 ; H_a:\mu \neq0$

Step-by-step explanation:

Consider the provided information.

Part (A)

A random sample of 100 students from a large university.

Increasing the sample size decreases the confidence intervals, as it increases the standard error.

If the researcher increase the sample size to 150 which is greater than 100 that will decrease the confidence intervals or the researcher could produce a narrower confidence interval.

Hence, the correct option is true.

Part (B)

The researcher wants to identify that whether there is any significant difference between the measurement of the blood pressure.

Therefore, the null and alternative hypothesis should be:

$H_o: \mu =0 ; H_a:\mu \neq0$

7 0

2 years ago