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BabaBlast [244]
3 years ago
13

Identity the pattern then write the next three terms in each sequence 512,256,128,64

Mathematics
2 answers:
Tomtit [17]3 years ago
8 0
Halting each time. 32, 16, 8
12345 [234]3 years ago
4 0

Answer:

32, 16, 8

Step-by-step explanation:

Pattern: divide by 2

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2 weeks after the encounter

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4 years ago
Tell me the answer of these question.
Nonamiya [84]
16. \ \ V= \pi r^2h \ \to \ r^2= \cfrac{V}{ \pi h}  \ \to \ \boxed{r= \sqrt{ \frac{V}{ \pi h} } } \\ \\ \\ 17.\ \ d= \frac{1}{2} at^2 \ \to \ 2d= at^2   \ \to \ t^2= \cfrac{2d}{a} \   \to \ \boxed{t= \sqrt{\frac{2d}{a}}} \\  \\   \\ 18. \ \ S=180(n-2) \ \to \  \cfrac{S}{180}=n-2 \ \to \ \boxed{n= \frac{S}{180}+2 }
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4 years ago
Which situation can be represented as a reflection?
kifflom [539]

Answer:

A chef flips a pancake over in a skillet

Step-by-step explanation:

5 0
3 years ago
A large university offers STEM (science, technology,engineenng. and mathematics) intemshups to women in STEM majors at the unive
ValentinkaMS [17]

Answer:

a) Probability of a randomly sampled women not being qualified for the internship = 0.223

b) Probability that at least 30 percent of the women in the sample will not meet the age requirement for the internships = 0.03216

c) A woman who does not meet the age requirement is more likely to be selected with a stratified random sample than with a simple random sample.

Step-by-step explanation:

Age | Probability

17 | 0.005

18 | 0.107

19 | 0.111

20 | 0.252

21 | 0.249

22 | 0.213

23 or older | 0.063

a) Only 20+ year olds are qualified for the internship

So, probability of being qualified for the internship = P(x ≥ 20)

Probability of not being qualified for the internship = P(x < 20) = P(x=17) + P(x=18) + P(x=19) = 0.005 + 0.107 + 0.111 = 0.223

b) According to the Central limit theorem, a sampling distribution of sample size as large as 100 selected from this population distribution will approximate a normal distribution. It also has that

Mean proportion of sampling distribution of women who do not meet the internship requirements = Population proportion of women who do not meet the internship requirements = p = 0.223

The standard deviation of the is given by

σₓ = √[p(1-p)/n]

n = sample size = 100

σₓ = √[(0.223×0.777)/100] = 0.041625833 = 0.04163

So, to obtain the probability that at least 30 percent of the women in the sample will not meet the age requirement for the internships

P(x ≥ 0.30)

We first standardize 0.30

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.30 - 0.223)/0.04163 = 1.85

The required probability

P(x ≥ 0.30) = P(z ≥ 1.85)

We'll use data from the normal probability table for these probabilities

P(x ≥ 0.30) = P(z ≥ 1.85) = 1 - P(z < 1.85)

= 1 - 0.96784 = 0.03216

c) Probability of women not meeting the internship requirements = 0.223

Probability of women meeting the internship requirements = 1 - 0.223 = 0.777

Or

Probability of women meeting the internship requirements = P(x ≥ 20)

= P(x=20) + P(x=21) + P(x=21) + P(x ≥ 23) = 0.777

But as the stratified sample only contains women who do not meet the internship requirements, it is more likely that A woman who does not meet the age requirement is selected with a stratified random sample than with a simple random sample.

Hope this Helps!!!

4 0
3 years ago
The following data points represent the number of attendees at each of the dance events hosted by Cloy
Viktor [21]
Put the numbers in order.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.
Step 2: Find the median.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.
Step 3: Place parentheses around the numbers above and below the median.
Not necessary statistically, but it makes Q1 and Q3 easier to spot.
(1, 2, 5, 6, 7), 9, (12, 15, 18, 19, 27).
Step 4: Find Q1 and Q3
Think of Q1 as a median in the lower half of the data and think of Q3 as a median for the upper half of data.
(1, 2, 5, 6, 7),  9, ( 12, 15, 18, 19, 27). Q1 = 5 and Q3 = 18.
Step 5: Subtract Q1 from Q3 to find the interquartile range.
18 – 5 = 13.

5 0
3 years ago
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