pls help me how do they get 9 7/5 = 10 2/5 ( its in the example number 1) plss help me i don't know :⊂

Find the equation of the tangent line to the curve y=x sqrtx at point (4,8)<br><br> Y=?

Viefleur [7K]

Answer: I think y is 32 im not good at this tho

Step-by-step explanation:

7 0

2 years ago

Sofia's taco truck lets customers choose 3 ingredients from the list below to add to their tacos. How many groups of 3 different

DIA [1.3K]

Using the combination formula, and considering that the list has 6 ingredients, it is found that 20 groups of 3 different ingredients can go on a taco.

The order in which the ingredients are chosen is not important, hence, the <em>combination formula</em> is used to solve this question.

<h3>Combination formula: </h3>

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem, 3 ingredients are chosen from a set of 6, hence:

$C_{6,3} = \frac{6!}{3!3!} = 20$

20 groups of 3 different ingredients can go on a taco.

To learn more about the combination formula, you can take a look at brainly.com/question/25990169

4 0

2 years ago

PLEASE HELP ME OUT ASAP! I WOULD REALLY APPRECIATE IT AND GOD BLESS

77julia77 [94]

To approximate the P(x<27) we need to find the z-score of the data, this will be given by:
z=(x-μ)/σ
where:
μ-mean
σ-standard deviation
x=27, μ=32, σ=4
z=(27-32)/4
z=-5/4
z=-1.25
thus
P(x<27)=P(z<-1.25)
=0.1056
=10.56%

Answer: 10.56%

5 0

3 years ago

What will be the area of the given shape? (please help)

Leni [432]

Answer:

620

Step-by-step explanation:

To find the area of strangely-shaped structures, like this one, it is easiest if we cut the shape into smaller, more recognizable, shapes..

We can first cut off the top/center rectangle of this shape , which has a width of 10, and a length of 12. The "area" is length × width.

So, we multiply 12 × 10 to find the area of this section.

12 × 10 = 120

if we look at the bottom half, we can see that there is an identical rectangle (because it has the same width of 10 and length of 12) ! So, we can simply add 120 to our total twice to count both of these areas.

(when putting areas of multiple portions together, we add--not multiply)

120 (first/top rectangle)

+ 120 (lower rectangle)

_______________

240

So, our area so far is 240.

Now, we need to find the horizontal (left-right) section of this shape.

We can see that it has a width of 10 (look at the right end), and a length of 38 14 + 10 + 14 (38 total)

So, to find the area of this portion, we will multiply its length × width, which we know is: 38 × 10

38 × 10 = 380.

So, we can now add our areas together:

240

+ 380

__________

620

So, our area of this shape is 620

6 0

1 year ago

Read 2 more answers

The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60

kodGreya [7K]

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

$n_1$ represent the sample after the preparation

$n_2$ represent the sample without preparation

$\bar x_1 =678$ represent the mean sample after preparation

$\bar x_2 =1837$ represent the mean sample without preparation

$s_1 =197$ represent the sample deviation after preparation

$s_2 =328$ represent the sample deviation without preparation

$\alpha=0.1$ represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}$ (1)

The point of estimate for $\mu_1 -\mu_2$

The appropiate degrees of freedom are $df=n_1+ n_2 -2$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)

The standard error is given by the following formula:

$SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}$

After replace in the formula for the confidence interval we got this:

$3.0 < \mu_1 -\mu_2$

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

5 0

2 years ago