7.23<br><br> What is two and<br> four tenths in<br> standard form?

What is the mode of the following set of data? 24, 31, 12, 38, 12, 15

BlackZzzverrR [31]

The mode is 12.

The mode is the value that appears most frequently in a data set.

6 0

3 years ago

I live 9 (rectangular) blocks from Tech, and I walk each morning. The trip is 3 blocks north and 6 blocks west; and I have to fo

Gnoma [55]

Answer:

We conclude that he have 84 possible complete trips to Tech.

Step-by-step explanation:

We know that he live 9 (rectangular) blocks from Tech. The trip is 3 blocks north and 6 blocks west. We calculate number of possible complete trips to Tech. We know that

$C^9_3=C^9_6$

We calculate

$C^9_{3}=\frac{9!}{3!(9-3)!}=\frac{9\cdot 8\cdot 7\cdot 6!}{3\cdot 2\cdot 1\cdot 6!} =84$

We conclude that he have 84 possible complete trips to Tech.

6 0

4 years ago

Select the statement that best justifies the conclusion based on the given information. l is in plane M, x is on line l Conclusi

egoroff_w [7]

Answer:

If a plane contains a line, it contains the points on the line

Step-by-step explanation:

I got it right on a quiz

8 0

3 years ago

If arc AB = 70°, how many degrees are in ∠ ADB?

tankabanditka [31]

If D is the center of the circle, ∠ADB = 70°, as that is where the arc gets its measure.

If D is on the circle and not on the arc, ∠ADB = 35°, half the measure of the arc's central angle.

If D is on arc AB, then ∠ADB = 145°, half the measure of the corresponding central angle.

If D is inside or outside the circle (and not one of the above), we need to know more about the measures of arcs intercepted by lines AD and BD.

5 0

3 years ago

A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. Th

igor_vitrenko [27]

Answer:

The 95% confidence interval for the mean breaking weight for this type cable is (767.47 lb, 777.13 lb).

Step-by-step explanation:

Our sample size is 41

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 41-1 = 40$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 40 and 0.025 in the two-sided t-distribution table, we have $T = 2.021$