All categories

3 years ago

Suppose $a$ and $b$ are the solutions to the quadratic equation $2x^2-3x-6=0$. Find the value of $(a+2)(b+2)$.

Mathematics

1 answer:

Charra [1.4K]3 years ago

4 0

We have that

2x²<span>-3x-6=0
using a graph tool----------> to resolve the second order equation
see the attached figure
the solution is
x=-1.137
x=2.637

therefore
a=-1.137
b=2.637
so
</span>(a+2)*(b+2)-----------> (-1.137+2)*(2.637+2)--------> 0.863*4.637---> 4

the answer is 4

You might be interested in

Not sure about this please help!!!

harina [27]

Answer:

i think its the last one

Step-by-step explanation:

as 15 - 3 is 12 and then the 7

hope this helps

i would appreciate it if u can heart and like my answer and maybe even give it 5 stars or brainliest pls i beg u thx !!! : )

8 0

3 years ago

Help <br> Brainleist<br> 15 points

geniusboy [140]

Z=22
180-120=60
2x+16=60
2x=44
Z=22

7 0

2 years ago

Read 2 more answers

A basketball player attemps 15 free throws in a game. He makes 9 of the attempted free throws. Which ratio describes the number

Ber [7]

Answer:

1) 3: 5

Step-by-step explanation:

Hope this helps!!! :)

5 0

3 years ago

I have a friend in need. Her name is Sienna and she has cancer along with co.vid. Please pray for her well being because she rea

Rzqust [24]

Answer:

I will

Step-by-step explanation:

7 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