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3 years ago

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Nora and her children went into a grocery store and she bought $5 worth of apples and bananas. Each apple costs $1 and each bana

na costs $0.50. She bought a total of 7 apples and bananas altogether. By following the steps below, determine the number of apples

1 answer:

qwelly [4]3 years ago

5 0

5 apples and 2 bananas

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Which statements about finding the area of the equilateral triangle are true? Check all that apply.

lara31 [8.8K]

Answer:

The correct options are 1, 2 and 4.

Step-by-step explanation:

The given triangle is an equilateral triangle. All the sides of an equilateral triangle are same. All interior angles are equal with measure 60 degree.

The distance from the center of an equilateral triangle to the midpoint of a side is known as apothem. In the given figure letter a represents the apothem.

Point D is the midpoint of BC,

$b=\frac{BC}{2}=\frac{8.7}{2}=4.35$

Use Pythagorean theorem in triangle COD. So, option 1 is correct.

$OC^2=OD^2+DC^2$

$5^2=a^2+(4.35)^2$

$25=a^2+18.9225$

$a=\pm 2.46526$

$a\approx\pm 2.5$

The distance can not be negative, so the length of the apothem is approximately 2.5 cm. Option 4 is correct.

The line OC bisects the angle C.

$tan(30^{\circ})=\frac{a}{b}$

$\frac{1}{\sqrt{3}=\frac{a}{4.35}$

$a\approx\pm 2.5$

Therefore option 2 is correct.

The perimeter of an equality triangle is

$P=3a=3(8.7)=26.1$

Option 3 is incorrect.

$A=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}(8.7)^2=32.77$

Option 5 is incorrect.

Therefore options 1, 2 and 4 are correct.

3 0

3 years ago

Read 2 more answers

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

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".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

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

$P(X>a)=0.1$ (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 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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3 years ago

How many points of intersection will the two lines below have? How do you know?

svp [43]

1, both functions are linear

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2 years ago

Which expression is equivalent to

andrezito [222]

Answer:

work it step by step on paper it really helps

Step-by-step explanation:

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Fraction equal to 1 that has 29 as a denominatior

d1i1m1o1n [39]

Answer:

29/29

Step-by-step explanation:

/ means division so 29/29=1

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3 years ago