I forgot how to solve equations such as 3x+5y=-5. Can you give me examples?

Which algebraic represent 15 less than x divided by 9

polet [3.4K]

X/9 - 15
X divided ( / ) by 9 minus ( - ) 9

6 0

2 years ago

GIVING OUT BRAINLIEST!! All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessari

Vsevolod [243]

Answer:

flour = 11.5, sugar has to be equal to 2.75 pounds.

Step-by-step explanation:

your 2 inequalities are:

2x + 3y <= 40

y <= 2x + 6

the second equation can be shown as:

-2x + y <= 6

solve the equality portion of these 2 inequalities.

those are the following equations:

2x + 3y = 40

-2x + y = 6

add these equations together and solve for y to get

y = 11.5

that appears to be a break even point.

when y = 11.5, the first inequality is solved as follows:

2x + 3y <= 40

replace y with 11.5 and the inequality becomes:

2x + 34.5 <= 40

solve for x to get

x <= 2.75

when y = 11.5, the second inequality is solved as follows:

-2x + y <= 6

replace y with 11.5 and the inequality becomes:

-2x + 11.5 <= 6

solve for x to get x >= 2.75

this indicates that, when y = 11.5, x has to be equal to 2.75.

both constraints are satisfied.

6 0

2 years ago

Read 2 more answers

Helppppppppppppppppp

bixtya [17]

The answer is C. Good luck

8 0

2 years ago

Read 2 more answers

Please help!! 20 POINTS

rjkz [21]

f(-3) would be 36.

When looking at synthetic division, the numbers across the top represent the coefficients of x^2, x and the constant in that order. Therefore, the equation is as follows.

2x^2 - 5x + 3

Now we can put -3 into the equation and solve.

2(-3)^2 - 5(-3) + 3

2(9) + 15 + 3

18 + 15 + 3

36

7 0

3 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago