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iren [92.7K]
3 years ago
14

Which set of data is most accurately represented by a logarithmic model?

Mathematics
1 answer:
Gnom [1K]3 years ago
7 0
Help I’ll mark you brainly!ununubyvybyvybybybybybybybybbybyubybybybbybybybbybyybybbyybbyububbububu
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If f(x) = 3x^2 express the value of f(2+h)-f(2)/h as a polynomial in terms of h.
LekaFEV [45]

Answer:

Step-by-step explanation:

8 0
2 years ago
CAN SOMEONE HELP PLEASE ASAP!!! ILL MAKE YOU BRAINLIEST
Anastaziya [24]

Answer:

I think the answer is D but I'm not 100 percent sure. I hope this helps

Step-by-step explanation:

I think I took this test before

8 0
3 years ago
What is the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34 Round yo
Galina-37 [17]

We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.

The probability of a data score between two z-scores is given by formula P(a.

Using above formula, we will get:

P(-1.32

Now we will use normal distribution table to find probability corresponding to both z-scores as:

P(-1.32

P(-1.32

Now we will convert 0.27351 into percentage as:

0.27351\times 100\%=27.351\%

Upon rounding to nearest tenth of percent, we will get:

27.351\%\approx 27.4\%

Therefore, our required probability is 27.4% and option C is the correct choice.

5 0
3 years ago
Mathematics
Vladimir [108]

Answer:

there no data

Step-by-step explanation:

Find the median of the first and third quartiles81, 69 ,90, 79, 59, 125, 95, 116, 183, 97, 92, 50, 122, 55, 9250, 55, 59, 69, 79, 81, 90,   92,     92, 95, 97, 116, 122, 125, 183.The first quartile: 50, 55, 59, 69, 79, 81, 90: median of the first quartile is 69The third quartile: 92, 95, 97, 116, 122, 125, 183: median of the third quartile is 116.The interquartile range is the difference between the median of both the first and third quartile:  which is 116 - 69 = 47.There are no outliers in this set of data

4 0
2 years ago
A recent broadcast of a television show had a 10 ​share, meaning that among 6000 monitored households with TV sets in​ use, 10​%
Kisachek [45]

Answer:

Null and alternative hypothesis

H_0: \pi \geq0.25\\\\H_1: \pi

Test statistic z=-26.82

P-value P=0

The null hypothesis is rejected.

It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

Step-by-step explanation:

We have to perform a hypothesis test of a proportion.

The claim is that less than 25% were tuned into the program, so we will state this null and alternative hypothesis:

H_0: \pi \geq0.25\\\\H_1: \pi

The signifiance level is 0.01.

The sample has a proportion p=0.1 and sample size of n=6000.

The standard deviation of the proportion, needed to calculate the test statistic, is:

\sigma_p=\sqrt{\frac{\pi(1-\pi)}{N} } =\sqrt{\frac{0.25(1-25)}{6000} } =0.0056

The test statistic is calculated as:

z=\frac{p-\pi+0.5/N}{\sigma_p} =\frac{0.1-0.25+0.5/6000}{0.0056}=\frac{-0.1499}{0.0056}  =-26.82

As this is a one-tailed test, the P-value is P(z<-26.82)=0. The P-value is smaller than the significance level (0.01), so the effect is significant.

Since the effect is significant, the null hypothesis is rejected. It is concluded that the proportion of households tuned into the program is less than 25%. The claim of the advertiser is rigth and got statistical support.

3 0
3 years ago
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