HELP PLEASE EXPLAIN

Mathematics

1 answer:

valkas [14]3 years ago

8 0

Answer:

57.1"

Step-by-step explanation:

i literally just measured my tv lol

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What fractions in this list are less than 1/2? 3/6,7/15,4/9,5/8

Vedmedyk [2.9K]

All we need to do is to find a common denominator for both 1/2 and the other numbers individually:

1/2 = 3/6 (so 3/6 is not less than 1/2 - both numbers are equal)
1/2 = 15/30 and 7/15 = 14/30 (so 7/15 is less than 1/2)
1/2 = 9/18 and 4/9 = 8/18 (so 4/9 is less than 1/2)
1/2 = 4/8 (so 5/8 is not less - it's greater than 1/2)

Answer: From the list above only 7/15 and 4/9 are less than 1/2.

7 0

4 years ago

Adequate Preparation for Retirement. In 2018, RAND Corporation researchers found that 71% of all individuals ages 66 to 69 are a

Svetllana [295]

Answer:

A - Hopes this helps.

Step-by-step explanation:

4 0

3 years ago

Simplify the expression using the distributive property. 6(3x + 4y)

Lyrx [107]

Answer:

18x+24y

Step-by-step explanation:

Distribute the 6 to the 3 (6x3), this gives you 18x. (Remember, keep the x.)

Now, distribute the 6 to the 4 (6x4), which gives you 24y.

Now write it as an equation, 18x+24y.

8 0

3 years ago

Read 2 more answers

Part of a university's professor's job is to publish his or her research. This task often entails reading a variety of journal a

goblinko [34]

Answer:

(18.1042, 35.2292) is a 90% confidence interval for the mean number of journal articles read monthly by professors.

Step-by-step explanation:

We have a small sample of size n = 12, $\bar{x} = 26.6667$ and s = 16.5163. A pivotal quantity for this case is given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which has a t distribution with n-1 degrees of freedom if we suppose that we take the sample from the normal population. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n - 1 = 12 - 1 = 11 degrees of freedom. As we want the 90% confidence interval, we have that $\alpha = 0.1$ and the confidence interval is $26.6667\pm t_{0.05}(\frac{16.5163}{\sqrt{12}})$ where $t_{0.05}$ is the 5th quantile of the t distribution with 11 df, i.e., $t_{0.05} = -1.7959$ . Then, we have $26.6667\pm (-1.7959)(\frac{16.5163}{\sqrt{12}})$ and the 90% confidence interval is given by (18.1042, 35.2292).