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3 years ago

Help me please I'm struggling

Mathematics

1 answer:

Umnica [9.8K]3 years ago

6 0

Answer:

Step-by-step explanation:

$-4\leq \frac{x+5}{3} \leq 7\\multiply~each~inequality~by~3\\-12\leq x+5\leq 21\\subtract~5~from~each~inequality\\-17 \leq x\leq 16$

cosθ+sinθ tan θ

=cosθ+sin θ×(sinθ)/(cosθ)

$=\frac{cos^2 \theta+sin ^2 \theta}{cos \theta} \\=\frac{1}{cos \theta} \\=sec \theta$

x²+(y-6)²=35

x²+y²-12y+36=35

x²+y²-12y+1=0

put x=r cosθ

y=r sin θ

square and add

x²+y²=r²cos²θ +r² sin²θ=r² (cos² θ+sin² θ)=r²

r²-12rsinθ+1=0

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Maurinko [17]

Answer:

A) 7, 2

B) 1, 5

C) 6, 3

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3 years ago

Read 2 more answers

Is 7b + 3 - b a polynomial?

Gemiola [76]

Answer:

yes!

Step-by-step explanation:

when we simplify this we will get 6b+3, which is a binomial so yes it is a polynomial!

hope this helps

5 0

3 years ago

Find a a pair of numbers of both 4y-4x=48 and y= x-11

stich3 [128]

Answer:

There is no solution because the slopes of the line are the same.

Step-by-step explanation:

4y - 4x = 48

+ 4x + 4x Add 4x to both sides

4y = 4x + 48 Divide both sides by 4

y = x + 12

Now set both equations equal to each other since they equal y

x + 12 = x - 11

-x - x Subtract x from both sides

12 ≠ -11 No Solution

If this answer is correct, please make me Brainliest!

4 0

3 years ago

Kyle solved 18 of 24 puzzles ina puzzle book. He says that he can use an equivalent fraction to find the percent of puzzles in t

Rashid [163]

$\frac{18}{24} = \frac{9}{12} = \frac{3}{4}$
The percentage is 75℅

8 0

3 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

3 years ago

Help me please I'm struggling ​

Help me please I'm struggling