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3 years ago

Consider the inequality x > - 1.5. determine whether each value of x makes this inequality true

Mathematics

1 answer:

musickatia [10]3 years ago

4 0

9514 1404 393

Answer:

no, yes, yes, no, yes

Step-by-step explanation:

You're looking for values greater than -1.5. All positive numbers will be greater than any negative number.

The two negative numbers among the answer choices are both to the left of -1.5 on the number line. They are less than -1.5, so are <em>not</em> greater than -1.5.

The negative numbers are No; the positive numbers are Yes.

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Only question numer 10.

leva [86]

Out of the six, only “u” wouldn’t work so the answer is d. 5

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4 0

3 years ago

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Find the increase in price of a single piece of design, when crafts design price is increased by 45% which makes a buyer to buy

Volgvan

Answer: Original price = $43.10, Increase = $19.40, Final price = $62.50

<u>Step-by-step explanation:</u>

Let x represent the original price, then

8x(1.45) < $500

$x$

x < $43.10

Increase is 0.45x

0.45($43.10)

= $19.40

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$43.10 + $19.40

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4 years ago

What is the value of 7 in 2,639,187

jolli1 [7]

The 7 should be one's place...?

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3 years ago

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(1 pt) Which answer best describes the solution to the equation 3m – 9 = 9 + 3m?

seropon [69]

3m-9=9+3m
add 9 to both sides
3m=18+3m
minus 3m both sides
0=18
false
no solution

the best discription would be an empty solution set

3 0

3 years ago

A random sample of 77 fields of durum wheat has a mean yield of 27.427.4 bushels per acre and standard deviation of 5.755.75 bus

xz_007 [3.2K]

Answer:

[ 21.79, 33.0 ]

Step-by-step explanation:

Given:

Sample size, n = 7

Sample mean, μ = 27.4

Standard deviation, σ = 5.75

Confidence level is 99%

also, population is normal i.e normally distributed

Thus,

Confidence interval = μ ± $z\frac{\sigma}{\sqrt n}$

For confidence level of 99%

z-value = 2.58

Therefore,

Lower limit of Confidence interval = μ - $2.58\times\frac{5.75}{\sqrt{7}}$

Lower limit of Confidence interval = 27.4 - $2.58\times\frac{5.75}{\sqrt{7}}$

Lower limit of Confidence interval = 21.79

Upper limit of Confidence interval = μ + $2.58\times\frac{5.75}{\sqrt{7}}$

Upper limit of Confidence interval = 27.4 + $2.58\times\frac{5.75}{\sqrt{7}}$

Upper limit of Confidence interval = 33.00

Hence, confidence interval = [ 21.79, 33.0 ]

5 0

3 years ago