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3 years ago

The lengths of the sides of a triangle are 3, 3, and the square root 3. Can the triangle be a right triangle?

2 answers:

8 0

First find the decimal equivalent of square root 3: SQRT(3) = 1.732 ( roughly)

If the base and height were each 3, then the hypotenuse would need to be:

3^2 + 3^2 = C^2
9 + 9 = C^2
18 = C^2
C = SQRT(18) = 4.24

This is larger than sqrt(3), so this cannot be a right triangle.

If one leg was 3 and the other leg was sqrt(3) then the hypotenuse would be:
3^2 + 1.73^2 = C^2
9 + 3 = C^2
12 = C^2
C = SQRT(12) = 3.46

This is larger than 3, this cannot be a right triangle.

The answer is b) no.

melisa1 [442]3 years ago

8 0

No
Because it does not obey Pythagoras theorem (a^2 + b^2 = c^2)

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Which of the following values would be an outlier in the following set? 1, 14, 17, 18, 19, 23

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1 would be the outlier

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3 years ago

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Find the measures of the legs of isosceles triangle ABC if AB= 2x+4,BC= 3x-1,AC=x+1 and the perimeter of the triangle ABC is 34

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(2x+4) +(3x-1) +(x+1)=34
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3 years ago

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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

2 years ago

(-0.3) ÷ (-0.2)+(-2.56) In whole numbers with steps please

Valentin [98]

Answer:

-1 to the nearest whole number.

Step-by-step explanation:

(-0.3) ÷ (-0.2)+(-2.56)

Order of operations - the division is done first:

= 1.5 = (-2.56)

= 1.5 - 2.56

= -1.06.

6 0

3 years ago

It takes Bruce 12 h to complete his training. It takes Oliver approximately 2 days, 2 h, and 15 min to complete

Stolb23 [73]

Answer:

Oliver completed his training work in 50.25 hours

Step-by-step explanation:

we know that

1 day= 24 hours

1 hour=60 minutes

we have

2 days, 2 h, and 15 min

Convert days to hours

2 days=2(24)=48 hours

Convert minutes to hour

15 min=15(1/60)=1/4=0.25 h

2 days, 2 h, and 15 min=(48)+(2)+(0.25)=50.25 h

therefore

Oliver completed his training work in 50.25 hours

3 0

3 years ago