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3 years ago

I need an answer ASAP.

Mathematics

1 answer:

melisa1 [442]3 years ago

7 0

6 positive tiles because there’s 6 negative so 6 positive will cancel it out

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If Hanna extends the pattern in the highlighted rows to include a ratio that has a first term of 20, what would be the second te

podryga [215]

Answer:

why do i see these so late

Step-by-step explanation:

5 0

4 years ago

Helppp please!!!! Thank you!!

vovangra [49]

Answer:first one A

Step-by-step explanation:

4 0

3 years ago

Plz help ill give u brainlist

g100num [7]

Step One
Find the base area of the large hexagon as though the smaller one was not removed.

Area = 3*Sqrt(3) * a^2 /2 where a is the length of one side of the hexagon
a = 5
Area = 3*sqrt(3) * 25/2 = 75 sqrt(3) / 2 of the large hexagon without the smaller one removed.

Step Two
Find the area of the smaller hexagon. In this case a = 4
Area2 = 3*sqrt(3)*16/2 = 3*sqrt(3)*8 = 24 sqrt(3)

Step Three
Find the area of the thick hexagonal area left by the removal of the small hexagon.

Area of the remaining piece = area of large hexagon - area of the small hexagon

Area of the remaining piece = 75 *sqrt(3)/2 - 24*sqrt(3)

Step Four
Find the volume of the results of the area from step 3
Volume = Area * h
h = 18
Volume = (75 * sqrt(3)/2 - 24*sqrt(3))* 18

I'm going to leave you with the job of changing all of this to a decimal answer. I get about 420 cm^3

3 0

3 years ago

If a hurricane was headed your way, would you evacuate? The headline of a press release issued January 21, 2009 by the survey re

Nikolay [14]

Answer:

The 98% confidence interval for population proportion of people who refuse evacuation is {0.30, 0.33].

Step-by-step explanation:

The sample drawn is of size, n = 5046.

As the sample size is large, i.e. n > 30, according to the Central limit theorem the sampling distribution of sample proportion will be normally distributed with mean $\hat p$ and standard deviation $\sqrt{\frac{\hat p (1-\hat p)}{n} }$ .

The mean is: $\hat p=0.31$

The confidence level (CL) = 98%

The confidence interval for single proportion is:

$CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]$

Here $z_{(\alpha /2)}$ = critical value and α = significance level.

Compute the value of α as follows:

$\alpha =1-CL\\=1-0.98\\=0.02$

For α = 0.02 the critical value can be computed from the z table.

Then the value of $z_{(\alpha /2)}$ is ± 2.33.

The 98% confidence interval for population proportion is:

$CI_{p}=[\hat p-z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} },\ \hat p+z_{(\alpha /2)}\times\sqrt{\frac{\hat p (1-\hat p)}{n} }]\\=[0.31-2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} },\ 0.31+2.33\times \sqrt{\frac{0.31\times(1-0.33)}{5046} } ]\\=[0.31-0.0152,\ 0.31+0.0152]\\=[0.2948,0.3252]\\\approx[0.30,\ 0.33]$

Thus, the 98% confidence interval [0.30, 0.33] implies that there is a 0.98 probability that the population proportion of people who refuse evacuation is between 0.30 and 0.33.

7 0

3 years ago

I have a 6 grader thst needs help with percentages; what assistance do you have here.

mr Goodwill [35]

Answer:

welp we can help them solve the question and we could also help you by teaching them with you if you need assistance

Step-by-step explanation:

5 0

3 years ago