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3 years ago

Classify the triangle by its angles and sides.

Mathematics

2 answers:

meriva3 years ago

7 0

Obtus is the right answer to this question

liraira [26]3 years ago

4 0

Answer:

obtuse scalene since 6 square plus 5 square is 36+25= 61, and 8 squared is 64, so 61<64, and if the legs are less than the hypotenuse, it would be an obtuse triangle. all the side are also unequal so it's a scalene.

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(-6)*(-6)*(-6)*z*z the expression is

ra1l [238]

Answer:

-216z^2

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

About how long is each side of a square that is 125 square miles

stiv31 [10]

Square = identical sides.
therefore, 125miles^2 = (square root 125)^2

square root 125 = 11.18033989miles

5 0

3 years ago

If M/DFC= 40 and M CD= 55 the BG= 25 80 135

Y_Kistochka [10]

Answer:

Step-by-step explanation:

GC and BD are 2 intersecting chords thus the measure of the angle formed is

∠DFC = 0.5( m arc CD + m arc BG ), that is

0.5(55 + BG) = 40 ( multiply both sides by 2 )

55 + BG = 80 ( subtract 55 from both sides )

BG = 25

6 0

3 years ago

Read 2 more answers

Bob is 4 years older than twice kim’s Age Paul is 1 year younger than 3 times kim’s Age Bob and Paul are the same age which equa

Art [367]

Answer:

4>1x

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day. Legacy

inessss [21]

Answer:

We conclude that the null hypothesis is rejected which means residents of Legacy Ranch use less water on average.

Step-by-step explanation:

We are given that the American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day.

Twenty-three owners responded, and the sample mean water use per day was 63 gallons with a standard deviation of 8.1 gallons per day.

Let $\mu$ = average water usage by residents of Legacy Ranch.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 67 gallons {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}

Alternate Hypothesis, $H_A$ : $\mu$ < 67 gallons {means that the residents of Legacy Ranch use less water on average}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean water use per day = 63 gallons

s = sample standard deviation = 8.1 gallons

n = sample of responded = 23

So, test statistics = $\frac{63-67}{\frac{8.1}{\sqrt{23} } }$ ~ $t_2_2$

= -2.368

The value of the test statistics is -2.368.

Now at 0.025 significance level, the t table gives critical value of -2.074 at 22 degree of freedom for left-tailed test. Since our test statistics is less than the critical values of t as -2.368 < -2.074, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the residents of Legacy Ranch use less water on average.

8 0

3 years ago