Solve and graph<br><br>26

I need yalls opinion on sumthin ok

melisa1 [442]

Answer:

?????......????.........??????

7 0

3 years ago

If yo are given PNM = MOP How would you prove that MNPQ is a parallelogram?

Fed [463]

Answer:

Response 1

Step-by-step explanation:

Response 1 is the only wording that is self-consistent. Response 1 is the appropriate response.

___

Response 2 argues about angles, then concludes sides are congruent. This makes no sense.

Response 3 identifies congruent segments, but they are not opposite sides. The conclusion is unsupported.

4 0

3 years ago

In ΔVWX, v = 390 inches, w = 390 inches and x=500 inches. What is the sum of ∠Vand∠W?

Stolb23 [73]

The answer would be 390+ 390 =780

6 0

3 years ago

The average expenditure on Valentine's Day was expected to be $100.89 (USA Today, February 13, 2006). Do male and female consume

AfilCa [17]

Answer:

Yes. The male and female consumers differ in the amounts they spend.

Step-by-step explanation:

We can express the null and alternative hypothesis as:

$H_0: \mu_m=\mu_w\\\\H_1: \mu_m\neq\mu_w$

It is assumed a significance level of 0.05.

The standard deviation of the difference of means is calculated as:

$s=\sqrt{\frac{s_m^2}{n_m} +\frac{s_w^2}{n_w} } =\sqrt{\frac{35^2}{40} +\frac{20^2}{30} } =\sqrt{30.625+13.333} =\sqrt{43.958} =6.63$

The test statistic is

$t=\frac{(M_m-M_w)-0}{s}=\frac{135.67-68.64}{6.63}=10.11$

The degrees of freedom are:

$df=n_1+n_2-2=40+30-2=68$

The P-value for t=10.11 is P=0, so it is smaller than the significance level. The null hypothesis is rejected.

We can conclude that male and female consumers differ in the amounts they spend.

5 0

4 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$