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3 years ago

5

If two triangles are similar, what is true regarding the sides of the two triangles? a. They are acute. c. They are proportional

. b. They are equal. d. They are similar.

2 answers:

Nastasia [14]3 years ago

8 0

C) they are proportional.

KonstantinChe [14]3 years ago

5 0

They arent simalr becuase they could be diffrent szes. Id say Proportional

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Give the slope-intercept form for the equation.<br><br> 5x+ 2y = -14

irga5000 [103]

Answer:

$y=-\frac{5}{2} x-7$

Step-by-step explanation:

$5x+2y=-14\\\frac{5x}{2} +\frac{2y}{2} =\frac{-14}{2}$

$\frac{5}{2}x+y=-7$

$y=-\frac{5}{2} x-7$

5 0

2 years ago

if it would take billy an hour to race over to his girlfriends house and two hours if he went 20 miles per hour slower how far a

mel-nik [20]

Recall your d = rt, distance = rate * time.

now, let's say she lives "d" miles away, and so if Billy goes say at "r" mph fast to her house, he gets there in 1hr.

now, if he goes 20 mph slower, namely " r - 20 " mph, he gets there in 2 hours, same "d" miles.

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{Billy}&d&r&1\\
\textit{slower Billy}&d&r-20&2
\end{array}
\\\\\\
\begin{cases}
d=r(1)\implies d=\boxed{r}\\
d=2(r-20)\\
----------\\
d=2\left( \boxed{d}-20 \right)
\end{cases}
\\\\\\
d=2d-40\implies 40=d$

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4 years ago

The pth percentile is a value such that approximately a. p percent of the observations are less than the value and (100 - p) per

kogti [31]

Answer:

p percent of the observations are less than the value and (100 - p) percent are more than this value.

Step-by-step explanation:

Given : The pth percentile is a value such that approximately

Solution :

Definition : A percentile is a measure used in statistics indicating the value below which a given percentage of observations in a group of observations falls

So, The pth percentile means p percentage of observations in a group of observations falls bellow the value

So, (100-p) percentage of observations in a group of observations falls above the value

So, Option a is true

Hence p percent of the observations are less than the value and (100 - p) percent are more than this value.

5 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

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 = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

Please help! files attached.

kotykmax [81]

The first question would be x + (x + 4). This is because Max has $x and Keisha has (x + 4) because she has four MORE THAN Max. I am pretty sure that the second one is A, but I am not positive. The first one is right though. I hope this helps!!!

5 0

3 years ago