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lions [1.4K]
3 years ago
10

Which of the following is a solution for the inequality 2x < 9?

Mathematics
1 answer:
Firdavs [7]3 years ago
8 0
The answer is x<9/2
hope it helps
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Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1
kondaur [170]
f will be increasing on the intervals where f'(x)>0 and decreasing wherever f'(x). Local extrema occur when f'(x)=0 and the sign of f'(x) changes to either side of that point.

f'(x) is positive when x is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that f(x) is increasing on the intervals [-4, -2] and (2, 4].

f'(x) is negative when x is between some number between -2 and -1, up to some number less than 2. So f(x) is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of f'(x) changes for some x between -2 and -1, and again to either side of x=2.
4 0
3 years ago
WHAT IS THE SQUARE ROOT OF 400
snow_tiger [21]
The square root of 400 is 20
7 0
3 years ago
Read 2 more answers
IM IN A TEST NOW WILL AWARD BRAINLIEST!!!!!
Oksanka [162]

The product has decreased by 80 %

<em><u>Solution:</u></em>

The product of the original numbers: xy

One number is decreased by 75 %

Let x be the number decreased by 75 %

new number = x - 75 % of x

new\ number = x - 75 \% \times x\\\\new\ number = x(1-75 \%)\\\\new\ number = x(1-\frac{75}{100}) = x(1-0.75) = 0.25x

Another number is decreased by 20 %

Let y be the number decreased by 20 %

new number = y - 20 % of y

new\ number = y - 20 \% y\\\\new\ number = y(1-20 \%)\\\\new\ number = y(1-\frac{20}{100}) = y(1-0.2) = 0.8y

<em><u>Now the product is given as:</u></em>

product = 0.25x \times 0.8y = 0.2xy

<em><u>Now we have to find the percent decreased</u></em>

Subtract new product and original product and then divide by original product . Multiply the result by 100 to get percentage

percent\ decrease = \frac{0.2xy - xy}{xy} \times 100\\\\percent\ decrease = \frac{-0.8xy}{xy} \times 100 = -0.8 \times 100 = -80

Here negative sign denotes decrease

Thus the product has decreased by 80 %

3 0
3 years ago
The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
MAXImum [283]

Answer:

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d.  The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

t(0.01)(49)= 1.299

t(0.05)= 1.677  

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0  for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test  we reject H0   as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ±  z∝/2  (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0
3 years ago
10
Archy [21]

Answer: 5h+9

Step-by-step explanation:

10h + (-5h)

5h+(6(+3))

6 0
2 years ago
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