Which is the value of the expression below?

5. Given the following triangle side lengths, identify the triangle as acute, right or obtuse. Show your work.

Roman55 [17]

Refer to the figure shown below.
We shall review each of the three given measurements and decide what type of triangle we have.

Measurement a.
a=3, b=4, c=5.
For a right triangle, c² = a² + b² (Pythagorean theorem)
a² + b² = 3² + 4² = 9 + 16 = 25
c² = 5² = 25
Answer:
This is a right triangle, because c² = a² + b².

Measurement b.
a=5, b=6, c=7.
For an acute triangle, c² < a² + b².
a² + b² = 5² + 6² = 25 + 36 = 61
c² = 7² = 49
Answer:
This is an acute triangle, because c² < a² + b².

Measurement c.
a=8, b=9, c=12.
For an obtuse triangle, c² > a² + b².
a² + b² = 8² + 9² = 64 + 81 = 145
c² = 12² = 144
Answer:
This is an acute triangle because c² < a² + b².

6 0

3 years ago

Read 2 more answers

If f(x)=x^2-11, for what values of z is f(x) < 25?

GuDViN [60]

Answer:

D

Step-by-step explanation:

f(x) < 25

x² - 11 < 25

x² < 36

-6 < x < 6

5 0

3 years ago

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

(c) The value of $P(\bar X\geq 69.1)$ is 0.3670.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$

And the standard deviation of the distribution of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74$

Thus, the distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b)

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

$=P(Z$

*Use a z-table for the probability.

Thus, the value of $P(\bar X$ is 0.7642.

(c)

Compute the value of $P(\bar X\geq 69.1)$ as follows:

Apply continuity correction as follows:

$P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)$

$=P(\bar X>69.6)$

$=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})$

$=P(Z>0.34)\\=1-P(Z$

Thus, the value of $P(\bar X\geq 69.1)$ is 0.3670.

7 0

3 years ago

Ameer biked 5 miles and then took a break. After the break,

PIT_PIT [208]

Answer:

C. You would be finding the total distance. You arent finding the rate because there is no division, and we dont know how long it took hime to ride the original 5. The answer is C.

3 0

2 years ago

PLEASE HELP!

ELEN [110]

Answer:

Colin's mean score = 17.10

The new mean is 18.00

Step-by-step explanation:

Given scores: 31, 10, 26, 16, 28, 31, 9, 11, 8, 1

number of terms: 10

mean is calculated by given formula = sum of all the data set/ number of terms in the data set.

mean for the 10 scores = (31+ 10+ 26+ 16+ 28+ 31+ 9+ 11+ 8+ 1 )/10

= 171/10 = 17.1

Colin's mean score = 17.10

____________________________________________

runs scored by Colin in 11th match = 27

number of terms: 11

new mean = new sum /number of terms

new mean = (old sum score + 27)/11 = (171+27)/11

new mean = 198/11 = 18

The new mean is 18.00

3 0

3 years ago