Find the volume of the cylinder.

How many triangles can be constructed with side lengths of 9.6 cm, 11.6 cm, and 21.2 cm?

nignag [31]

Answer: A) 0 triangles

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Explanation:

Adding up the two smaller sides gets us 9.6+11.6 = 21.2, but this result is not larger than the third side of 21.2

For a triangle to be possible, we need to be able to add any two sides and have the sum be larger than the third remaining side. This is the triangle inequality theorem.

I recommend you cutting out slips of paper with these side lengths and trying it out yourself. You'll find that a triangle cannot be formed. The 9.6 cm and the 11.6 cm sides will combine to form a straight line that is 21.2 cm, but a triangle won't form.

As another example of a triangle that can't be formed is a triangle with sides of 3 cm, 5 cm, and 8 cm. The 3 and 5 cm sides add to 3+5 = 8 cm, but this does not exceed the third side. The best we can do is form a straight line but that's not a triangle.

In short, zero triangles can be formed with the given side lengths of 9.6 cm, 11.6 cm, and 21.2 cm

3 0

3 years ago

Translate the sentence into an equation: the sum of a number divided by 7 and 9 equals 3

ludmilkaskok [199]

X/7 + 9 = 3
a number: x
a number divided by 7: x/7
sum of number divided by 7 and 9: x/7 + 9 (and means plus)
equals 3: x/7 + 9 = 3

3 0

3 years ago

What is the answer to this algebra problem

Sergeu [11.5K]

The problem said : -10,8 =-2 and -15,9=-6

5 0

4 years ago

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A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec

Ket [755]

Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

$Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18$

The sample size is, n = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

$\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191$

(a)

The null hypothesis is:

H₀: The usual load is 13 credits, i.e. μ = 13.

Assume that the significance level of the test is, α = 0.05.

Construct a (1 - α) % confidence interval for population mean to check the claim.

The (1 - α) % confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\times SE$

For 5% level of significance the two tailed critical value of z is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct the 95% confidence interval as follows:

$CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)$

As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

$P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z$

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

3 0

3 years ago

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Nina had 9 apples to share with 5 of her friends. If all 6 people get equal share , how many apples will each person get? How do

Umnica [9.8K]

Nina + her 5 friends = 6 people
Then she has 9 apple to share with her and her friends.
Let’s calculate how many apple will each people get when they equally divide it with each other:
=> 9 – 6 = 3
Thus already have 1 apple each, now, let’s divide the remaining 3 apples
=> 3 / 6 = .5
So they have half of apple each. Let’s add
=> 1 + .5 = 1.5
Each person will get 1.5 apple.
=> 1.5 * 6 = 9

7 0

3 years ago

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