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3 years ago

Graph the solution set for this inequality -6x-3y<-18

Mathematics

2 answers:

Vikentia [17]3 years ago

7 0

Answer:

Hope It Helps

Yuliya22 [10]3 years ago

5 0

Answer:

it is 27 because of my cacculations

Step-by-step explanation:

you would fist subtract your numbers and thne get rid of the 5

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Twice the sum of -4 and a number is the same as the number decreased by 7/2

White raven [17]

2.(-4 + x) = x - 7/2

-8 + 2x = x - 7/2

2x - x = -7/2 + 8

x = 9/2

3 0

3 years ago

Anne can make 6 headbands 3/5 from meters of cloth. Which expression shows the amount of cloth needed to make 1 headband? What i

Marina CMI [18]

The amount of cloth needed to make 1 headband is 10 meters.

<h3>Unit value</h3>

Number of headbands Anne made = 6

Total clothes used = 3/5 meters

Cloth needed to make 1 headband = Number of headbands Anne made / Total clothes used

= 6 ÷ 3/5

= 6 × 5/3

= (6 × 5) / 3

= 30/3

= 10 meters

Therefore, the amount of cloth needed to make 1 headband is 10 meters.

Learn more about unit value:

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4 0

2 years ago

You and two friends visit a pizza shop. There are 15 toppings. 5 toppings are meat and the rest are vegetable. Each of you order

natta225 [31]

I think 25 %i could be wrong though

3 0

3 years ago

Expand and simplify completely (2x+3y)^4

Solnce55 [7]

Answer:

16x^4+96x^3y+216x^2 y^2+216xy^3+81y^4

3 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago