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sukhopar [10]
3 years ago
15

Find the value of x x<-1 and x>1

Mathematics
1 answer:
vodka [1.7K]3 years ago
7 0

Answer:2

Step-by-step explanation: pretty simple, trust me :)

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No clue how to do this someone please help
Free_Kalibri [48]

Answer:

I recommend trying this it is real tutors that explain to you on how to do it it's free

7 0
3 years ago
4 + (11 - 3)^2 simplified answer fast pls
ValentinkaMS [17]

Answer:

68

Step-by-step explanation:

4 + (11 - 3)^2

4 + 8^2

4 + 64

= 68

3 0
3 years ago
What is 67.50 with 5 percent sales tax
Sindrei [870]
3.375 is your answer
6 0
3 years ago
Read 2 more answers
The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated
cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

<em>Let </em>p_1<em> = proportion of cockroaches that died on glass surface.</em>

<em />p_2<em> = proportion of cockroaches that died on plasterboard surface.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of cockroaches that died on glass surface = \frac{18}{36} = 0.50

\hat p_2 = sample proportion of cockroaches that died on plasterboard surface = \frac{25}{36} = 0.694

n_1 = sample of cockroaches on glass surface = 36

n_2 = sample of cockroaches on plasterboard surface = 36

So, <u><em>test statistics</em></u>  =  \frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36}  } }

                               =  -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0
3 years ago
Shreya and beth each improved their yards by planting rose bushes and ornamental grass. They brought their supplies from the sam
frutty [35]

Answer:

  • <u><em>The cost of one rose bush is $ 9 and the cost on one bunch pf ornamental grass is $ 8.</em></u>

Explanation:

Name the variables and build a system of two equations with two unknowns.

<u>1. Variable r</u>: cost of one rose bush

<u>2. Variable g</u>: cost of one bunch of ornamental grass.

<u>3. First equation</u>:

  • Shreya spent $68 on 4 rose bushes and 4 bunches of ornamental grass.

        4r + 4g = 68

<u>4. Second equation</u>:

  • Beth spent $115 on 11 rose bushes and 2 bunches of ornaments grass.

  • 11r + 2g = 115

<u>5. System of equations</u>:

  • 4r + 4 g = 68     equation 1
  • 11r + 2g = 115     equation 2

<u>6. Solve the system</u>

a) Divide the equation 1 by 2:

  • 2r + 2g  =  34     equation 1a
  • 11r + 2g = 115     equation 2

b) Subtract equation 1a from equation 2:

  • 11r - 2r = 115 - 34
  • 9r = 81
  • r = 81 / 9
  • r = 9

c) Substitute the value of r into the equation 1a:

  • 2(9) + 2g = 34
  • 9 + g = 17
  • g = 17 - 9
  • g = 8

<u>7. Conclusion</u>:

The cost of one rose bush is $ 9 and the cost on one bunch pf ornamental grass is $ 8.

7 0
3 years ago
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