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andriy [413]
1 year ago
14

While pushing a shovel into the ground with a force of 555 newtons ant an angle of 44 ∘ 44 ∘ to the ground. Find the magnitudes

of the horizontal and vertical components of the force to the nearest whole newton.
Mathematics
1 answer:
Nataly [62]1 year ago
3 0

The Horizontal component is 554.889N and the vertical component is

9.8235 Newton

<h3>Resolution of Forces</h3>

Given Data

  • Force = 555 Newton
  • Angle of Applied force = 44°
  • Fx = Horizontal component of the applied force
  • Fy = the vertical component of the applied force

Resolving the vertical force

Fy = 555 sin44°

Fy = 555 *0.01770

Fy = 9.8235 Newton

Resolving the horizontal force

Fx = 555 cos 44°

Fx = 555 *0.9998

Fx = 554.889 Newton

Learn more about forces here:

brainly.com/question/25997968

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It is estimated that 30% of households own a riding lawn mower. A sample of 17 households is studied. What is the standard devia
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Answer:

1.89 is the standard deviation of households who own a riding mower

Step-by-step explanation:

Let the random variable X denote the number of households owning a riding lawn mower. From the given information, X will be a Binomial random variable with parameters; n = 17 and  p = 30% = 0.3

The standard deviation of X;

\sqrt{n*p*(1-p)}

\sqrt{17*0.3*(1-0.3)} =1.89

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2 years ago
From some petroleum, you can get 30% of paraffin. How much petroleum do you need in order to get 18.75 tons of paraffin?
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5 0
3 years ago
Not sure how I would solve this
Simora [160]
<h3>Answer:   -6/5</h3>

Explanation:

The blue diagonal line goes through the two points (0,2) and (5,-4). These are shown as the dark blue enlarged points. You can pick any other points you want that are on the diagonal line, though these are the easiest as they stand out the most.

Use the slope formula to find the slope through these points

m = (y2-y1)/(x2-x1)

m = (-4-2)/(5-0)

m = (-6)/(5)

m = -6/5

The negative slope means the line goes downhill as you move from left to right along the diagonal line.

7 0
3 years ago
A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken
mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

X=1 number of citrus trees that are infected with the citrus red mite.

n=5 random sample taken

\hat p=\frac{1}{5}=0.2 estimated proportion of citrus trees that are infected with the citrus red mite.

p true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values would be given by:

t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.2(1-0.2)}{(\frac{0.01}{1.96})^2}=6146.56  

And rounded up we have that n=6147

5 0
3 years ago
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