Find the approximate height of the flagpole.

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

B the answer is 60 feet

DochEvi [55]3 years ago

7 0

Tan(40)=Height÷50

Height
=50(tan(40))
=42 feet (to nearest whole number)

Hence answer is C

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Calculate the unit rate: write your answer as a decimal 1/2 mile in ever 1/3 hour

Alexandra [31]

Answer:

$unit \: rate = \frac{distance}{time}$

unit rate is similar to speed / velocity:

$unit \: rate = \frac{ (\frac{1}{2} )}{ (\frac{1}{3})} \\ \\ = \frac{1}{2} \div \frac{1}{3} \\ \\ = \frac{1}{2} \times \frac{3}{1} \\ \\ = > { \boxed{ \boxed{unit \: rate = \frac{3}{2} \: miles \: per \: hour}}}$

[unit rate must have units including per <time> ]

7 0

3 years ago

The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen

faltersainse [42]

Answer:

$a=150.7 -0.385*10.2=146.773$

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Solution to the problem

Let X the random variable that represent the scores on the LSAT of a population, and for this case we know the distribution for X is given by:

$X \sim N(150.7,10.2)$

Where $\mu=150.7$ and $\sigma=10.2$

We want to find a value a, such that we satisfy this condition:

$P(X>a)=0.65$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65

If we use condition (b) from previous we have this:

$P(X$