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2 years ago

10

Please help me !! is/is notcan/cannotis/is notcan/cannot

1 answer:

zlopas [31]2 years ago

7 0

Answer:

isn't

can't

Step-by-step explanation:

hjjhfyuhvdgujbcff

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A test preparation claims that more than 50% of the students who take their test prep course improve their scores by at least 1

satela [25.4K]

Answer:

Step-by-step explanation:

Corresponding scores before and after taking the course form matched pairs.

The data for the test are the differences between the scores before and after taking the course.

μd = scores before taking the course minus scores before taking the course.

a) For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

b) We would assume a significance level of 0.05. The P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.

Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.

c) The mean difference for the sample scores is greater than or equal to zero

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2 years ago

Pls help i will mark brainleist

lara [203]

Answer:

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3 years ago

Find the area of the figure<br>a) 114cm2<br>b) 350cm2<br>c) 418cm2<br>d) 447cm2<br>

uysha [10]

Answer: I thin it's A because you just multiply 19*6

Step-by-step explanation:

4 0

3 years ago

Ms Hart had 40 dollars to spend on costume boxes for the musical. If she bought 12 boxes and received 4 dollars in change, how m

artcher [175]

Answer:each box cost =$3

Step-by-step explanation:

Amount of money Ms Hart has to spent on costume boxes=$40

Amount gotten as change after purchase=$4

Amount actually used up buying costume boxes =$40-$4=36

Since she bought 12 boxes , each box cost =$36/ 12=$3

5 0

2 years ago

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago