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fiasKO [112]
2 years ago
15

Which sets of ordered pairs show equivalent ratios? use the grid to help you. check all that apply

Mathematics
1 answer:
ss7ja [257]2 years ago
5 0
1 ,3,last one this is the answer for your Question that’s all
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Solve for z -9(d+z)=-2z+59
neonofarm [45]
-9(d+z)=-2z+59\\\\\text{Use multiply distributive property:}\\a\cdot(b+c)=ab+ac\\\\-9d-9z=-2z+59\ \ \ |+9d\\\\-9z=-2z+59+9d\ \ \ |+2z\\\\-7z=59+9d\ \ \ \ |:(-7)\\\\\boxed{z=-\dfrac{59+9d}{7}}

7 0
3 years ago
Read 2 more answers
Kindlyy solve this question..
Bas_tet [7]

Answer:

Volume = 21688.37 in.³

Step-by-step explanation:

Volume of a sphere : V=4/3πr³

4/3π17.3³ ≈ 21688.37025

hope this helps :D

3 0
3 years ago
According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For
SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

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.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

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

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0
2 years ago
What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?
yKpoI14uk [10]
a_1=8
a_2=a_1+d=8+d
a_3=a_2+d=(8+d)+d=8+2d
a_4=a_3+d=(8+d)+2d=8+3d
...
a_9=8+8d=48\implies d=5
a_{10}=8+9d
...
a_{23}=8+22d=118
8 0
3 years ago
Read 2 more answers
To find the next number in the sequence: 6, 9, 12, 15, of the following? you would need to do which of the following.
ycow [4]
C. add 3 to 15 because the sequence is +3 every time
8 0
2 years ago
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