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Marysya12 [62]
2 years ago
7

Use the box-and-whisker plots to determine which statements are true.

Mathematics
1 answer:
ExtremeBDS [4]2 years ago
5 0

The true statements are :

the average monthly temperatures of New Orleans are not as varied as that of Indianapolis.

The median of the average monthly temperatures of New Orleans is equal to the upper quartile of the average monthly temperature of Indianapolis.

<h3>What are the true options?</h3>

A box plot is used to study the distribution and level of a set of scores. The box plot consists whiskers. The whiskers represents the minimum and maximum scores.

On the box, the first line to the left represents the lower (first) quartile. The next line on the box represents the median.  The third line on the box represents the upper (third) quartile.

Range for New Orleans = 85 - 50 = 35

Range for Indianapolis = 75 - 25 = 50

Median for New Orleans = 70

Upper quartile for  Indianapolis = 70

To learn more about box plots, please check: brainly.com/question/27215146

#SPJ1

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Oksi-84 [34.3K]

Answer:

     A

B    D    C          graph of the triangle

statment/reasons

segment AD is an angle bicestor of angleBAC, AD is the hight of the triangle/Given

angleBAD=angleCAD/Def. of angle bicestor

AD=AD/Reflexive Property

angleADB=angleADC/Def. of Right Angle

triangleABD=triangleCAD/ASA

angleABD=angleACD/CPCTC

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4 0
3 years ago
A regular polygon has its interior angles all equal to 174°
Vika [28.1K]
The shape has sixty sides

working out:
180 - 174 = 6
360 / 6 = 60
5 0
3 years ago
The average speed of a car driving down freeway is 70mph with a standard deviation of 5mph. Thid speed is normally distributed.
ikadub [295]

Answer:

0.081,0.2621,0.645,0.919

Step-by-step explanation:

Let X be the average speed of a car driving down freeway

Given that X is normal with mean as 70mph with a standard deviation of 5mph.

To convert to Z score we do

Z =\frac{x-70}{5}

a)  the probability that a randomly selected car is driving more than 77mph

=P(X>77)\\=P(Z>1.4)\\=0.081

b) probability that a randomly selected car is driving between 65 and 69 mph

= P(-1

c) the probability that a randonly selected car is driving between 67 and 77mph

=P(-0.6

d) the probability that a randomly selected car is driving less than 77mph

=P(Z

5 0
3 years ago
An example problem in a Statistics textbook asked to find the probability of dying when making a skydiving jump.
MArishka [77]

Answer:

(a) 0.999664

(b) 15052

Step-by-step explanation:

From the given data of recent years,  there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths.

So, the probability of death is \frac{21}{3000000}==0.000007.

Assuming, this probability holds true for each skydiving and does not change in the present time.

So, as every skydiving is an independent event having a fixed probability of dying and there are only two possibilities, the diver will either die or survive, so, all skydiving can be regarded as is Bernoulli's trial.

Denoting the probability of dying in a single jump by q.

q=7\times 10^{-6}=0.000007.

So, the probability of survive, p=1-q

\Rightarrow p=1-7\times 10^{-6}=0.999993.

(a) The total number of jump he made, n=48

Using Bernoulli's equation, the probability of surviving in exactly 48 jumps (r=48) out of 48 jumps (n=48) is

=\binom(n,r)p^rq^{n-r}

=\binom(48,48)(0.999993)^{48}(0.000007)^{48-48}

=(0.999993)^{48}=0.999664 (approx)

So, the probability of survive in 48 skydiving is 0.999664,

(b) The given probability of surviving =90%=0.9

Let, total n skydiving jumps required to meet the surviving probability of 0.9.

So, By using Bernoulli's equation,

0.9=\binom {n }{r} p^rq^{n-r}

Here, r=n.

\Rightarrow 0.9=\binom{n}{n}p^nq^{n-n}

\Rightarrow 0.9=p^n

\Rightarrow 0.9=(0.999993)^n

\Rightarrow \ln(0.9)=n\ln(0.999993) [ taking \log_e both sides]

\Rightarrow n=\frac {\ln(0.9)}{\ln(0.999993)}

\Rightarrow n=15051.45

The number of diving cant be a fractional value, so bound it to the upper integral value.

Hence, the total number of skydiving required to meet the 90% probability of surviving is 15052.

3 0
3 years ago
Suppose you have pennies, nickels, and dimes in your pocket. If you pull out four coins, what possible amount of money could you
mina [271]
Well first we need to know how many pennies, nickles, and dimes there are.
3 0
3 years ago
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