If data set a has a larger standard deviation than data set b, data set a is less spread out than data set b

The given statement is false.

Step-by-step explanation:

Given that there are two data sets A and B. data set A has a larger std deviation

Standard deviation is nothing but the square root of variance

Variance is the sum of squares of deviations of each entry from the mean.

Hence larger the std deviation larger is the spread

Since data set A has a larger standard deviation, the statement

Data set A is less spread out than Data set B is false.

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)

180-142=38 for the acute angle next to the entry

90+38=128

180-128=52

a=52

step-by-step explanation: