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3 years ago

Can someone help me with this math problem?

Mathematics

1 answer:

Ainat [17]3 years ago

6 0

Answer:

$x=7,\\m\angle LKN = 84^{\circ}$

Step-by-step explanation:

1. For KM to be the angle bisector of angle LKN, it must be bisect LKN into two equal angles. Therefore, we have the following equation:

$9x-21=6x,\\-21=-3x,\\x=\boxed{7}$

2. Substituting $x=7$ :

$\angle LKN=9x-21+6x=9(7)-21+6(7)=63-21+42=\boxed{84^{\circ}}$

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Rasek [7]

Answer:

Step-by-step explanation:

We use BODMAS/BIDMAS here.

First, we do 5 × 2, as multiplication comes before subtracting and adding.

5 × 2 = 10

Now we can just add and subtract.

5 - 10 = -5

-5 + 11 = 6

3 0

3 years ago

A glacier in Alaska moves about 29.99 a day about how much further will a move in a thousand days and it will move in a hundred

Lelu [443]

Try this: (29.99 units/day)(1000 days) = 29990 units.

Could you possibly include the given units of measurement in this post?

5 0

3 years ago

Help with this math question please

kvv77 [185]

First question.

We will build up the 3-dimensional object from the dimensions shown in the view above. In doing this, the object is as shown in figure 1. So, we need to find the volume of this object. Let's cut off this object into two objects, so in doing this we can find the whole volume as the sum of the volume in green plus the volume in red (both are cubes), so:

$V_{t} = V_{g}+V_{r} = 8 \times 3 \times 5 + 8 \times 5 \times 8 = 440 ft^{3}$

So the answer is G.

Second question

The answer is D. Let's analyze it:

Given that r represents the radius of the pizza and n the number of slices, so if we substitute r in this equation, then the values for n are:

$n = r^{2}+1$

$n=2^{2}+1=5$
$n=4^{2}+1=17$
$n=5^{2}+1=16$
$n=8^{2}+1=65$
$n=10^{2}+1=101$

These are the values of n shown in the table above.

6 0

4 years ago

Glenna says that the height of her house is 1 yard is her claim reasonable

muminat

No, her answer is not reasonable because one yard is equal to 3 feet.

4 0

3 years ago

Read 2 more answers

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago