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2 years ago

Can someone please help me with this math problem. It is rate of change.

Mathematics

1 answer:

Shalnov [3]2 years ago

8 0

Answer:

Step-by-step explanation:

I think 2700miles /(4+2/3) equal 578,5

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10x+18<-2<br>help me please!!!

sergey [27]

10x ∠-20
x∠-20/10

x∠-2

this is your answer
hope it helps

3 0

3 years ago

Read 2 more answers

NEED HELP ASAP!!!!!!

tino4ka555 [31]

Answer:

x=57

Step-by-step explanation:

63+60= 123

A triangle is 180 degrees

180-123= 57

x=57

5 0

3 years ago

Read 2 more answers

"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen

Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± $Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ ]

Where

p' is the sample proportion/point estimator of the population proportion

$Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ is the margin of error (d) of the confidence interval.

$Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648$

$d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$

$d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }$

$(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}$

$n*(d*Z_{1-\alpha /2})^2= p'(1-p')$

$n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}$

$n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}$

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0

3 years ago

Determine whether the situation is linear or nonlinear, and proportional or nonproportional

Arlecino [84]

Answer:

linear and non-proportional

Step-by-step explanation:

When the same amount is added each week, the relationship is linear. When the initial value is 50, not zero, the relationship is non-proportional.

6 0

3 years ago

Question number 8. Prove that 3 + √5 is an irrational number.

Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

$\\ \sf \: 3 + \sqrt{5} = \frac{a}{b} \\ \\ \qquad \: \tiny \sf{(where \: \: a \: \: and \: \: b \: \: are \: \: integers \: \: and \: \: b \: \neq \: 0)} \\$

$\\ \sf \: \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \\$

Since, a, b and 3 are integers. So,

$\\ \sf \: \frac{p - 3b}{b} \\ \\ \qquad \tiny \sf{ \: (is \: \: a \: \: rational \: \: number \:) } \\$

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

$\\$

Hence, 3 + √5 is an irrational number.

7 0

3 years ago