Milo is volunteering at a food pantry. In the first hour, he packs 3 boxes every

Dylan owes half as much money as he used to owe. If he used to owe $28, which of the following expressions would reflect how muc

IrinaVladis [17]

Answer:

it is the second one i had this question. mark as brainliest

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Aflati x numar natural nul

Advocard [28]

Exceseuor if im wrong sorry

5 0

3 years ago

Which equation can be solved by using this system of equations? startlayout enlarged left-brace 1st row y = 3 x superscript 5 ba

Mariulka [41]

An equation which can be used to solve the given system of equations is 3x⁵ - 5x³ + 2x² - 10x + 4 = 4x⁴ + 6x³ - 11.

<u>Given the following data:</u>

y = 3x⁵ - 5x³ + 2x² - 10x + 4

y = 4x⁴ + 6x³ - 11

<h3>What is a system of equations?</h3>

A system of equations can be defined an algebraic equation that only has two (2) variables and can be solved simultaneoulsy.

Equating the given equations, we have:

y = y

3x⁵ - 5x³ + 2x² - 10x + 4 = 4x⁴ + 6x³ - 11

3x⁵ - 5x³ + 2x² - 10x + 4 - (4x⁴ + 6x³ - 11) = 0

3x⁵ - 5x³ + 2x² - 10x + 4 - 4x⁴ - 6x³ + 11 = 0

3x⁵ - 4x⁴ - 11x³ + 2x² - 10x + 15 = 0

Read more on equations here: brainly.com/question/13170908

3 0

2 years ago

A past survey of students taking a standardized test revealed that % of the students were planning on studying engineering in c

Angelina_Jolie [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval is $-0.00870$

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 1068000$

The first proportion $\r p_1 = 0.084$

The second sample size is $n_2 = 1476000$

The second proportion is $\r p_2 = 0.092$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = (100 - 95)\%$

$\alpha = 0.05$

From the normal distribution table we obtain the critical value of $\frac{ \alpha }{2}$ the value is

$Z_{\frac{\alpha }{2} } =z_c= 1.96$

Now using the formula from the question to construct the 95% confidence interval we have

$(\r p_1 - \r p_2 )- z_c \sqrt{ \frac{\r p_1 \r q_1 }{n_1} + \frac{\r p_2 \r q_2 }{n_2} }$

Here $\r q_1 = 1 - \r p_1$

=> $\r q_1 = 1 - 0.084$

=> $\r q = 0.916$

and

$\r q_2 = 1 - \r p_2$

=> $\r q_2 = 1 - 0.092$

=> $\r q_2 = 0.908$

So

$(0.084 - 0.092 )- (1.96)* \sqrt{ \frac{0.092* 0.916 }{1068000} + \frac{0.084* 0.908 }{1476000} }$

$-0.00870$

3 0

3 years ago

Evaluate the following integral using trigonometric substitution.

wariber [46]

Answer:

Step-by-step explanation:

1. Given the integral function $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as $asin \theta$ i.e $x = a sin\theta$ .

All integrals in the form $\int\limits {\sqrt{a^{2} -x^{2} } } \, dx$ are always evaluated using the substitute given where 'a' is any constant.

From the given integral, $\int\limits {7\sqrt{49-x^{2} } } \, dx = \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx$ where a = 7 in this case.

The substitute will therefore be $x = 7 sin\theta$

2.) Given $x = 7 sin\theta$

$\frac{dx}{d \theta} = 7cos \theta$

cross multiplying

$dx = 7cos\theta d\theta$

3.) Rewriting the given integral using the substiution will result into;

$\int\limits {7\sqrt{49-x^{2} } } \, dx \\= \int\limits {7\sqrt{7^{2} -x^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -(7sin\theta)^{2} } } \, dx\\= \int\limits {7\sqrt{7^{2} -49sin^{2}\theta } } \, dx\\= \int\limits {7\sqrt{49(1-sin^{2}\theta)} } } \, dx\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, dx\\since\ dx = 7cos\theta d\theta\\= \int\limits {7\sqrt{49(cos^{2}\theta)} } } \, 7cos\theta d\theta\\= \int\limits {7\{7(cos\theta)} }}} \, 7cos\theta d\theta\\$

$= \int\limits343 cos^{2} \theta \, d\theta$

8 0

4 years ago