Word problem. Help (:

Graph the following system of linear inequalities. Identify at least two points in the solution: y < 5 - 2x | x + 5y > -7

romanna [79]

Answer:

(1,2) and (2,-1)

Step-by-step explanation:

we have

$y< 5-2x$ ----> inequality A

The solution of the inequality A is the shaded area below the dashed line $y=5-2x$

$x+5y>-7$ ---->inequality B

The solution of the inequality B is the shaded area above the dashed line $x+5y=-7$

The solution of the system of inequalities is the triangular shaded area between the two dashed lines

If a ordered pair is a solution of the system of inequalities, then the ordered pair must lie on the shaded area

Two points in the solution are(1,2) and (2,-1)see the attached figure

4 0

4 years ago

What is the slope of the line?

Burka [1]

Answer:

ur the one who stole my 50 points last time, now i steal urs, i remember ur profile picture. but its 2/3

7 0

3 years ago

Makayla has $8 to buy tickets at the school fair. Each ticket costs @1.5. Which inequality best represents how many tickets she

cricket20 [7]

Answer:

$n\le 5$

Step-by-step explanation:

Let n be the number of tickets Makaya has to buy.

If the cost of one ticket is $1.5, then the cost of n tickets is $1.5n.

Makayla has $8 to buy tickets at the school fair, thus

$1.5n\le 8\\ \\n\le \dfrac{8}{1.5}\\ \\n\le \dfrac{80}{15}\\ \\n\le \dfrac{16}{3}\\ \\n\le 5\dfrac{2}{3}$

The maximum number of tickets Makaya can buy is 5, so

$n\le 5$

5 0

4 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 83 type K batteries and a sample of 77

agasfer [191]

Answer:

1. Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1\neq \mu_2$ {mean voltage for these two types of

batteries is different]

2. Test Statistics value = -5.06

4. Decision for the hypothesis test is that we will reject null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, $\mu_1$ = true mean voltage for type K batteries.

$\mu_2$ = true mean voltage for type Q batteries.

So, Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1\neq \mu_2$ {mean voltage for these two types of

batteries is different]

The test statistics we use here will be :

$\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ follows $t_n__1+n_2-2$

where, $X_1bar$ = 9.29 and $X_2bar$ = 9.65

$s_1$ = 0.374 and $s_2$ = 0.518

$n_1$ = 83 and $n_2$ = 77

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2} }{83+77-2} }$ = 0.45 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77} } }$ follows $t_1_5_8$

= -5.06

At 0.05 or 5% level of significance t table gives a critical value between (-1.98,-1.96) to (1.98,1.96) at 158 degree of freedom. Since our test statistics is less than the critical table value of t as -5.06 < (-1.98,-1.96) so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that mean voltage for these two types of batteries is different.

7 0

4 years ago

What’s is the answer to this equation -3 - 5n = 1 - 6n

Tomtit [17]

-3 - 5n = 1 - 6n

+6n +6n

-3 + n = 1

+3 +3

n = 4

Hope this helps! ;)

4 0

3 years ago