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3 years ago

13

I can quite figure out how to go about this problem l, we are writing equations by defining a variable, writing an equation, the

n solving

1 answer:

galben [10]3 years ago

3 0

Answer:

14 ft

Step-by-step explanation:

Let x represent the number of segments of the divided wood beam

Let y represent the total length of the wood beam.

Given that each segment measures 3.5, the equation to express this scenario is given as

y = 3.5x

Thus, to find the length of the wood beam, substitute x = 4 into the equation.

We would have:

y = 3.5(4)

y = 14

Length of the wood beam = 14 ft

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The effect of a monetary incentive on performance on a cognitive task was investigated. The researcher predicted that greater mo

riadik2000 [5.3K]

Answer:

1) $H_0:\mu_5=\mu_{25}=\mu_{50}$

2) $H_a:\mu_{50}>\mu_{25}>\mu_{5}$

3) A Type I error happens when we reject a null hypothesis that is true. In this case, that would mean that the conclusion is that there is evidence to support the claim that the greater the incentive, the more puzzles are solved, but that in reality there is no significant difference.

4) A Type II error happens when a false null hypothesis is failed to be rejected. In this case, that would mean that there is no enough evidence to support the claim that the greater the incentive, the more puzzles are solved, but in fact this is true.

5) The probability of a Type I error is equal to the significance level, as this is the chance of having a sample result that will make the null hypothesis be rejected.

Step-by-step explanation:

As the claim is that the greater the incentive, the more puzzles were solved, the null hypothesis will state that this claim is not true. That is that there is no significant relation between the incentive and the amount of puzzles that are solved. In other words, the mean amount of puzzles solved for the different incentives is equal (or not significantly different):

$H_0:\mu_5=\mu_{25}=\mu_{50}$

The research (or alternative hypothesis) is that the greater the incentive, the more puzzles were solved. That means that the mean puzzles solved for an incentive of 50 cents is significantly higher than the mean mean puzzles solved for an incentive of 25 cents and this is significantly higher than the mean puzzles solved for an incentive of 5 cents.

$H_a:\mu_{50}>\mu_{25}>\mu_{5}$

A Type I error happens when we reject a null hypothesis that is true. In this case, that would mean that the conclusion is that there is evidence to support the claim that the greater the incentive, the more puzzles are solved, but that in reality there is no significant difference.

A Type II error happens when a false null hypothesis is failed to be rejected. In this case, that would mean that there is no enough evidence to support the claim that the greater the incentive, the more puzzles are solved, but in fact this is true.

The probability of a Type I error is equal to the significance level, as this is the chance of having a sample result that will make the null hypothesis be rejected.

4 0

3 years ago

work out an estimate for the number of boxes of grass seed Helen needs.to find your estimate, round each of the numbers in your

Dima020 [189]

Answer:

A circle is a bounded object with points from its center to its circumference is equidistant.

Area of a circle = πr²

Where :

π = pi = 3.14

R = radius = diameter / 2 = 20m / 2 = 10m

3.14 x 10² = 314m²

Number of boxes needed to cover the garden = area of the garden / coverage of one box

314m² / 58m² = 5.14

5 boxes to one significant figure

<h2>Brainliest pls</h2>

3 0

2 years ago

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{2})^{2}+cos^{2}(x)=1$

$cos^{2}(x)=1-\frac{1}{4}$

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

$sin(y)=\frac{\sqrt{2}}{2}$

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

3 years ago

7(28) = 7(20 + 8) explanation

forsale [732]

Step-by-step explanation:

7(28)=196

7(20)=140 + 7(8)=56

140+56=196

196=196

6 0

3 years ago

Read 2 more answers

X^3+7x^2+13x+4=0 Show all work

Softa [21]

$\displaystyle\\
x^3+7x^2+13x+4=0\\\\
x^3+\underbrace{4x^2+3x^2}_{7x^2}+\underbrace{12x+x}_{13x}+4=0\\\\
(x^3+4x^2) + (3x^2 + 12x)+(x+4)=0\\\\
x^2(x+4)+3x(x+4)+(x+4)=0\\\\
(x+4)(x^2 + 3x +1)=0\\\\
x+4 = 0~~~\Longrightarrow~~~\boxed{x_1 = -4}\\\\
x^2 + 3x +1=0\\\\
x_{23}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{-3\pm \sqrt{3^2-4\cdot 1 \cdot 1} }{2\cdot 1}= \frac{-3\pm \sqrt{5} }{2}\\\\
\boxed{x_2 = \frac{-3-\sqrt{5} }{2}}\\\\
\boxed{x_3 = \frac{-3+\sqrt{5} }{2}}$

4 0

3 years ago