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Verdich [7]
3 years ago
14

Which graph represents a function?

Mathematics
2 answers:
frozen [14]3 years ago
6 0

Answer:

the answer is the top right one

Solnce55 [7]3 years ago
3 0
Top right one ! is the answer
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Help me with C number please. Show me how did you do that. It’s due in 30 minutes.
Paha777 [63]
I’m not 100% sure but isn’t it x2 +xy +y2?
6 0
2 years ago
Read 2 more answers
Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.
zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

  31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

__

b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

  6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

  = 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

__

c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

_____

The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

7 0
3 years ago
Pleaseee helppppp!!!
maria [59]

Answer:

Addition property of equality

Step-by-step explanation:

3/4 was added to each side

4 0
3 years ago
a sample of 546 boys aged 6–11 was weighed, and it was determined that 89 of them were overweight. A sample of 508 girls aged 6–
AleksAgata [21]

Answer:

p-value:  0.6527

Step-by-step explanation:

Hello!

You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:

Sample 1 (Boys aged 6-11)

n₁= 546

x₁= 89

^p₁= x₁/n₁ = 89/546 ≅0.16

Sample 2 (girls aged 6-11)

n=508

x= 74

^p= x/n = 74/508 ≅ 0.15

If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:

H₀: ρ₁ = ρ₂

H₁: ρ₁ ≠ ρ₂

This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:

Z=<u>      (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂)        </u> ≈ N(0;1)

    √(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))

ρ`=<u> x₁ + x₂  </u> = <u>  89+74     </u> = 0.154 ≅ 0.15

     n₁ + n₂     546 + 508

Z⁰ᵇ =<u>          (0.16-0.15) - (0)                    </u>

       √(0.15 * (1 - 0.15) * (1/546 + 1/508))

Z⁰ᵇ = 0.45

I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:

P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527

p-value:  0.6527

Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05

Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.

I hope it helps!

3 0
3 years ago
Anna’s cell phone plan charges her $30 per month plus a $150 one-time activation fee. Evelyn’s cell phone plan charges her $20 p
AveGali [126]

Given:

Anna’s cell phone plan charges her $30 per month plus a $150 one-time activation fee.

Evelyn’s cell phone plan charges her $20 per month, plus a $450 one-time activation fee.

To find:

The number of months after which the costs for the girls’ cell phone plans the same.

Solution:

Let x be the number of months.

Total cost = Fixed cost + Variable cost

According to the question, cost equation for Anna’s cell phone is

y=150+30x           ...(i)

Cost equation for Evelyn's cell phone is

y=450+20x         ...(ii)

Equate (i) and (ii) to find the time after which the costs for the girls’ cell phone plans the same.

150+30x=450+20x

30x-20x=450-150

10x=300

Divide both sides by 10.

x=\dfrac{300}{10}

x=30

Therefore, the costs for the girls’ cell phone plans the same after 10 months.

5 0
3 years ago
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