Factor completely: 4d^3+3d^2-14d

Answers

Answer 1

Answer: I got d(4d2+3d−14) hope it help

Related Questions

Which of the below is the quotient for the expression 1/5 divided by 4?A) 1/5B) 4/5C) 1/20D) 20
Enter the sum of numbers as a product of their GCF. 45 + 30
PLEASE HELP RLLY NEED IT GUYS
"The municipal transportation authority determined that 58% of all drivers were speeding along a busy street. In an attempt to reduce this percentage, the city put up electronic speed monitors so that drivers would be warned if they were driving over the speed limit. A follow-up study is now planned to see if the speed monitors were effective. The null and alternative hypotheses of the test are H0 : π = 0.58 versus Ha : π < 0.58. It is planned to use a sample of 150 drivers taken at random times of the week and the test will be conducted at the 5% significance level. (a) What is the most number of drivers in the sample that can be speeding and still have them conclude that the alternative hypothesis is true? (b) Suppose the true value of π is 0.52. What is the power of the test? (c) How could the researchers modify the test in order to increase its power without increasing the probability of a Type I Error?"
72 is 45% of what number

The design of a concrete mix requires 2,314 lb/yd3 of gravel having a moisture content of 3.5% and absorption of 4.2%, and 899 lb/yd3 of sand having a moisture content of 5.7% and absorption of 1.4%, and 244 lb/yd3 of free water. What is the weight of the mixing water per cubic yard that should be used at the job site?

Answers

Answer:

Weight of mixing water=224.541 lb

Step-by-step explanation:

Taking 1 cubic yard of concrete

Mass of gravel = 2314 lb

Moisture content = 3.5% Absorption 4.2%

Extra water needed = (4.2-3.5)*2314/100= 16.198 lb

Mass of sand= 899 lb

Moisture content = 5.7% Absorption =1.4%

Water released = (5.7-1.4)*899/100= 38.657 lb

Free water = 244 lb

Weight of mixing water = free water + extra water needed-water released = 244+16.198-38.657=224.541 lb

Weight of mixing water=224.541 lb

* 9. Two fifths of what number is 120?
2/5

Answers

Answer:

300

Step-by-step explanation:

120 ÷ 2/5 = 300

Of = multiply
Let X represent the number we are looking for
2/5 x X=120
2/5X=120
Divide both sides by 2/5
X= 300
L

An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value for the given sample size and confidence level.(a) n = 6, CL = 90%(b) n = 21, CL = 98%(c) n = 29, CL = 95%(d) n = 12, CL = 99%

Answers

(a) For n = 6, CL = 90%,

The degrees of freedom: 5, Critical t-value: 2.571

(b) For n = 21, CL = 98%,

The degrees of freedom: 20, Critical t-value: 2.845

The degrees of freedom: 28, Critical t-value: 2.048

(d) For n = 12, CL = 99%,

The degrees of freedom: 11, Critical t-value: 3.106

Use the concept of critical t- value defined as:

A critical value is a number that is used in hypothesis testing to compare to a test statistic and evaluate whether or not the null hypothesis should be rejected. The null hypothesis cannot be rejected if the test statistic's value is less extreme than the crucial value.

(a) Given that,

n = 6 and a confidence level of 90%,

The degrees of freedom are,

n-1 = 6-1

The degrees of freedom = 5.

To find the critical t-value,

Look it up in the t-distribution table using a confidence level of 90% and a degree of freedom of 5.

From the table,

The critical t-value is approximately 2.571.

(b) Given that,

n = 21 and a confidence level of 98%,

The degrees of freedom are,

n-1 = 21-1

The degrees of freedom = 20.

By referring to the t-distribution table with a confidence level of 98% and degrees of freedom of 20,

The critical t-value is approximately 2.845.

n = 29 and a confidence level of 95%,

The degrees of freedom are,

n-1 = 29-1

The degrees of freedom = 28

Using the t-distribution table with a confidence level of 95% and degrees of freedom of 28,

The critical t-value is approximately 2.048.

(d) Given that,

n = 12 and a confidence level of 99%,

The degrees of freedom are,

n-1 = 12-1

The degrees of freedom = 11

By consulting the t-distribution table with a confidence level of 99% and degrees of freedom of 11,

The critical t-value is approximately 3.106.

To learn more about statistics visit:

brainly.com/question/30765535

#SPJ12

Final answer:

To find the degrees of freedom and critical t-value for each given sample size and confidence level, we can use the t-distribution and a t-table. The degrees of freedom (df) for each sample is equal to the sample size minus 1. The critical t-value can be found using the t-table with the corresponding degrees of freedom and the confidence level.

Explanation:

To find the degrees of freedom and critical t-value for each given sample size and confidence level, we can use the t-distribution and a t-table. The degrees of freedom (df) for each sample is equal to the sample size minus 1. For example, for (a) n = 6, df = 6 - 1 = 5. The critical t-value can be found using the t-table with the corresponding degrees of freedom and the confidence level.

For (a) n = 6, CL = 90%, the critical t-value is approximately 1.943.

For (b) n = 21, CL = 98%, the critical t-value is approximately 2.861.

For (c) n = 29, CL = 95%, the critical t-value is approximately 2.045.

For (d) n = 12, CL = 99%, the critical t-value is approximately 3.106.