A direct variation includes the points (21,7) and (27,Y. Find<br>value of y.<br>

Please help if u can! greatly appreciated<br> don't answer if u dont know PLEASE

fenix001 [56]

a. log₈8 = 1

b. log₇6/5

c. log₉3¹

<h3>The solution to the problem are</h3>

1. log₈ 2 * 4 would produce an integer

= log₈8 = 1

The output is an integer

2. log₇6/5 would give us an irrational number

= log₇1.2

This would give us an irrational number

3. log₉3¹ would give us a rational number

= log₉9^ $^\frac{1}{2}$ = 1/2 log⁹9

= 1/2

Read more on rational numbers here: brainly.com/question/12088221

#SPJ1

6 0

2 years ago

You solve a brain teaser 0.75 of the time. describe the likeihood

d1i1m1o1n [39]

.75 is the same as 75 %, therefore the likelihood of you solving the brain teaser is 75% of the time. Hope that helped.

3 0

3 years ago

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally dist

Fofino [41]

Answer:

a) $8469-1.645\frac{100}{\sqrt{64}}=8448.4$

$8469+1.645\frac{100}{\sqrt{64}}=8489.6$

So on this case the 90% confidence interval would be given by (8448.4;8489.6)

b) $8469-2.054\frac{100}{\sqrt{64}}=8443.33$

$8469+2.054\frac{100}{\sqrt{64}}=8494.68$

So on this case the 96% confidence interval would be given by (8443.33;8494.68)

Step-by-step explanation:

Assuming this complete problem: On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with s = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of s.

(a) Assuming this to be the case, if a sample of 64 modified bars resulted in a sample average yield point of 8469 lb, compute a 90% CI for the true average yield point of the modified bar. (Round your answers to one decimal place.)

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=8469$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=100$ represent the population standard deviation

n=64 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.645$

Now we have everything in order to replace into formula (1):

$8469-1.645\frac{100}{\sqrt{64}}=8448.4$

$8469+1.645\frac{100}{\sqrt{64}}=8489.6$

So on this case the 90% confidence interval would be given by (8448.4;8489.6)

(b) How would you modify the interval in part (a) to obtain a confidence level of 96%? (Round your answer to two decimal places.)

Since the Confidence is 0.96 or 96%, the value of $\alpha=0.04$ and $\alpha/2 =0.02$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.02,0,1)".And we see that $z_{\alpha/2}=2.054$

Now we have everything in order to replace into formula (1):

$8469-2.054\frac{100}{\sqrt{64}}=8443.33$

$8469+2.054\frac{100}{\sqrt{64}}=8494.68$

So on this case the 96% confidence interval would be given by (8443.33;8494.68)

7 0

3 years ago

A group of students each are given a certain number of green and yellow marbles

allochka39001 [22]

Answer: A,d,f

Step-by-step explanation:

Because you have to simplify the ratios of the colors.

6 0

3 years ago

A nutritionist attempts to determine an association between where food is prepared and the number of calories the food contains.

Oksi-84 [34.3K]

Given the conditional relative frequency table below which was generated by column using frequency table data comparing the number of calories in a meal to whether the meal was prepared at home or at a restaurant.

Number of Calories and Location of Meal Preparation.
Home Restaurant Total
≥ 500 calories 0.15 0.55 0.28

< 500 calories 0.85 0.45 0.72

Total 1.0 1.0 1.0

To determine whether there is an association between where food is prepared and the number of carories the food contain, we recall that an "association" exists between two categorical variables if the column conditional relative frequencies are different for the columns of the table. The bigger the differences in the conditional relative frequencies, the stronger the association between the variables. If the conditional relative frequencies are nearly equal for all categories, there may be no association between the variables. Such variables are said to be <span>independent.

For the given conditional relative freduency, we can see that there is a significant difference between the columns of the table.
i.e. 0.15 is significantly different from 0.55 and 0.85 is significantly different from 0.45

Therefore, we can conclude from the given answer options that t</span><span>here is an association because the value 0.15 is not similar to the value 0.55</span>

8 0

3 years ago

A direct variation includes the points (21,7) and (27,Y. Findvalue of y.​

A direct variation includes the points (21,7) and (27,Y. Findvalue of y.