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3 years ago

Please help I failed this test five times.

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

3 0

Step-by-step explanation:

a 10kg box putting pushed with a force of 24 Newtons and a fraction of 12N. Determine coefficient of Friction and Determine acceleration of the Box

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The sum of the squares of two consecutive even integers is 884. determine the integers.

Digiron [165]

Answer:

(2x+1)(1) = 84. 2x+1= 84. 2x = 83.

Step-by-step explanation:

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3 years ago

Convert 180 inches to feet.

Crazy boy [7]

Answer:

180"/12 = 15 ft

Step-by-step explanation:

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3 years ago

Read 2 more answers

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago

Decide whether the change is an increase or a decrease, and find the percent change. Original number = 35; New number = 16.... P

Elodia [21]

Answer: 60 decrease or 150 decrease

Step-by-step explanation:

3 0

3 years ago

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such tha

natita [175]

Answer:

Step-by-step explanation:

Given that:

Population Mean = 7.1

sample size = 24

Sample mean = 7.3

Standard deviation = 1.0

Level of significance = 0.025

The null hypothesis:

$H_o: \mu = 7.1$

The alternative hypothesis:

$H_a: \mu > 7.1$

This test is right-tailed.

$degree \ of \ freedom= n - 1 \\ \\ degree \ of \ freedom = 24 - 1 \\ \\ degree \ of \ freedom = 23$

Rejection region: at ∝ = 0.025 and df of 23, the critical value of the right-tailed test $t_c = 2.069$

The test statistics can be computed as:

$t = \dfrac{ \hat X - \mu_o}{\dfrac{s}{\sqrt{n}}}$

$t = \dfrac{ 7.3-7.1}{\dfrac{1}{\sqrt{24}}}$

$t = \dfrac{0.2}{0.204}$

t = 0.980

Decision rule:

Since the calculated value of t is lesser than, i.e t = 0.980 < $t_c = 2.069$ , then we do not reject the null hypothesis.

Conclusion:

We conclude that there is insufficient evidence to claim that the population mean is greater than 7.1 at 0.025 level of significance.

6 0

3 years ago