-6 -5(6x-4) Explain please

Mathematics

1 answer:

Zolol [24]2 years ago

7 0

$\qquad\qquad\huge\underline{{\sf Answer}}$

let's evaluate ~

$\qquad \sf \dashrightarrow \: - 6 - 5(6x - 4)$

$\qquad \sf \dashrightarrow \: - 6 - [(5 \cdot6x) - (5 \cdot4)]$

$\qquad \sf \dashrightarrow \: - 6 - (30x - 20)$

$\qquad \sf \dashrightarrow \: - 6 -30x + 20$

$\qquad \sf \dashrightarrow \: - 30x + 14$

I hope you understood the whole procedure ~

You might be interested in

A coin is tossed 10 times. Find the chance of getting exactly 5 heads. Find the chance of obtaining between four and six heads i

Korolek [52]

The chance of getting 5 heads is 50% because there are only two sides to a coin hope this helps

4 0

3 years ago

Find the slope of each of the following

Degger [83]

Answer:

m = ⅔

m = 2

m = ⅖

m = -8/5

7 0

3 years ago

38 students were asked to report how many siblings they have - DATA: 8 has 0 siblings; 13 had 1 siblings; 12 had 2 siblings; 3 h

olga_2 [115]

8 out of 38 percentage is 21% chance

6 0

3 years ago

Read 2 more answers

??? Plz answer this!!!!!

lianna [129]

Answer:

So the main objective is to find how many students per bus, and we see that 6 students went by car, so we subtract 6 to 360, which would be 354. Now you do 354/6 because there are 6 buses and 354 students in total, which should give you 59.

Hope that answers your question

4 0

4 years ago

An F5 tornado is classified as having mean wind speeds greater than 261 mph. In an effort to classify a recent tornado, 20 wind

mina [271]

Complete Question

An F5 tornado is classified as having mean wind speeds greater than 261 mph. In an effort to classify a recent tornado, 20 wind speed measurements were randomly selected from within the path of the tornado. Suppose the mean wind speed for this tornado really is greater than 261 mph with a sample standard Deviation 41.09 mph. Based on the data, if we fined little to no evidence against the null hypothesis that the mean wind speed is 261 mph, we have made _______________.

Answer:

$P-value=0.3322$