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4 years ago

Which two tables represent the same function?

Mathematics

1 answer:

Natalka [10]4 years ago

4 0

Can you make it more clear to read

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in the gym one sixth of the students are 5th graders one third are fourth graders and one fourth of the remaining students are 3

Sati [7]

Dear Sunshine76, first add 1/6+1/4+1/3=3/4. Then, find 3/4 of 96 to get 24 students in 3rd grade.

5 0

3 years ago

The process of producing pain-reliever tablets yields tablets with varying amounts of the active ingredient. It is claimed that

lisabon 2012 [21]

Answer:

p = 0.0116

Step-by-step explanation:

We are given;

Population mean; μ = 200 mg

Sample mean; x¯ = 194.3 mg

Standard deviation; σ = 21 mg

Sample size; n = 70

Formula for z-score is;

z = (x¯ - μ)/(σ/(√n))

z = (194.3 - 200)/(21/√70)

z = -5.7/2.51

z = -2.27

From the z-score table attached, the p-value at z = -2.27 is p = 0.0116

3 0

3 years ago

Question 1 In which expression should the exponents be multiplied? (1 over 5) to the power of 2 (1 over 5) to the power of 6

BlackZzzverrR [31]

Answer:

No, should be added

Step-by-step explanation:

The question is:

$(1/5)^2(1/5)^6 =$
$(1/5)^{2+6} =$
$(1/5)^8$

Use the rule:

$a^ba^c= a^{b+c}$

6 0

3 years ago

Read 2 more answers

The image is not set to scale btw but need the answer

Len [333]

<A = 62 and <ABE = 18

Sum of interior angles in a triangle = 180

<E + <A + <ABE = 180

<E + 62 + 18 = 180

<E + 80 = 180

<E = 100

Answer: first option

100

6 0

4 years ago

Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=10

Tju [1.3M]

Answer: 0.8238

Step-by-step explanation:

Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with $\mu=106$ and $\sigma=15$ .

Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.

Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-

$P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]$

Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238

4 0

3 years ago