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bulgar [2K]
3 years ago
11

Find the gradient of the curve y=(2x-3)^5 at the point (2,1)

Mathematics
2 answers:
Setler [38]3 years ago
6 0
<h3><em>Hope I helped you!</em></h3><h3><em>Success!</em></h3>

AfilCa [17]3 years ago
5 0

Answer:

(-10,1)

Step-by-step explanation:

Caculating the gradient:

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HELPP!!! 30 POINTS!!! WILL MARK BRAINLYIST!! Step 1: Written Response (30 points) Using complete sentences, compare the fat cont
yan [13]

Answer:

median a= 65 b= 75 c=68

This shows the median of each fat content of each restaurant showing that restaurant c is closer to that of 'restaurant a ' than a is to 'restaurant b' and near to 1/7 lower than that of restaurant b which shows they are all highly skewed with each other upon their graph.

The measures of spread start with 55 to 72 for 'restaurant a' where the upper and lower quartiles  are 70-60 showing a distribution 10-17 into the whisker plot.

Compared to restaurant b which shows a measure of spread start with 65 to 90 for 'restaurant b' where the upper and lower quartiles are 85 - 68 into the whisker plot and show a spread of 3 in the left exterior and 5 spread in the right exterior to the upper quartile. So in comparison to restaurant a that restaurant b has a greater spread in the interquartile range = 15  where restaurant a =10.

For the outer quartile the comparison to restaurant a is twice as small meaning restaurant a is greater by over double and shown as   53 <a < 60 which when showing both outer quartiles we can use the amounts being 7 and 70<a<72 = 2

and show closed dots on equality line number line separately showing a<-7  and a>2 for each outer exterior quartile. For restaurant b this shows opposite similar equalities 65<b< 68 = b>-3 and   82<b<89 = b>7

So the differences are smaller for b for left side by 4 and larger for b by 5 up on the right side.

We compare both to restaurant c and find c =   60<c<61 = c>-1 and 70<c<71 = c>1 so the differences are that the outer quartile for c through the other quartiles = -2> c < -6 . This shows how smaller c is compared to a and b outer quartiles). We can also prove that while the outer quartiles are much more smaller for c than a and b we cna prove that c actually has an inner quartile more similar to c and closer distribution of b as the median is more closer for a and c where b has a greater output for median as restaurant b has the higher fat content and greater distribution within the inner quartiles over all.

Summary findings Restaurant c outer quartile is moderately skewed as they show -1 -0.5  on the left side and 1-0.5 on the right side. Restaurant a has a closer median inner quartile to c and closer distribution of inner distribution of c. It's output outer quartile distribiution is a distribution that is higher than b but a smaller inner quartile compared to b, when this happens then the distribution spread shows less range and fewer products to account for.

So i think restaurant b has the higher fat content to its menu.

Where restaurant a must be the healthiest as it holds the lowest range and larger gap is such range for healthier food.ie) when compared to the others. 

Meanings of what we are asked.

In a box and whisker plot: the ends of the box are the upper and lower quartiles, so the box spans the interquartile range. the median is marked by a vertical line inside the box. the whiskers are the two lines outside the box that extend to the highest and lowest observations.

Skewness refers to distortion or asymmetry in a symmetrical bell curve, or normal distribution, in a set of data. If the curve is shifted to the left or to the right, it is said to be skewed. Skewness can be quantified as a representation of the extent to which a given distribution varies from a normal distribution.

As a general rule of thumb:

If skewness is less than -1 or greater than 1, the distribution is highly skewed.

If skewness is between -1 and -0.5 or between 0.5 and 1, the distribution is moderately skewed.

If skewness is between -0.5 and 0.5, the distribution is approximately symmetric.

4 0
3 years ago
Which is bigger 6.951 or 6.9313?
fomenos

Answer:

6.51

Step-by-step explanation:


7 0
3 years ago
Two questions! PLease help!! I need it (:
Arlecino [84]
The answer to the horizontal method is
4n³+2n²-6n-9

The answer to the FOIL problem is
10x²-25x-35

Hope this helps!
7 0
3 years ago
I need help on number 18
zloy xaker [14]

Answer:

Volume = 35185.8 km^{3}

Step-by-step explanation:

Recall the formula for the volume of a cone: Volume = \pi  R^{2}  \frac{h}{3}

Where R is the radius of the cone's circular base (in your case half of 80 km,(that is 40 km)

And h stands for the height of the cone (in your case 21 km)

We replace such in the formula and calculate its final value:

Volume = \pi 40^{2} \frac{21}{3} = \pi  1600 \frac{21}{3} = \pi 1600 (7) = 35185.83772...

We round it to the number of decimals your problem request (let's say to one decimal) and also include the cubic units:

Volume = 35185.8 km^{3}

7 0
3 years ago
Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
Kobotan [32]
<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

5 0
3 years ago
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