0.87 written as a fraction is _____. How would you work it out?

Mathematics

2 answers:

tatyana61 [14]3 years ago

7 0

1. 0.87 is the same as 87/100
87/100

Alexxx [7]3 years ago

4 0

Move the decimal all the way to the right. It can go to the right two times so put the 87 under 100. 87/100. (You put as many zeros in the denominator based on how many times you move the decimal over. Ex: 10, 100, 1,000)

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(21.23) A study of the effect of exposure to color (red or blue) on the ability to solve puzzles used 42 subjects. Half the subj

Cloud [144]

Answer:

There is enough statistical evidence to say that the means are different.

Step-by-step explanation:

The question is incomplete:

"The two-sample t statistic for comparing the population means has value (±0.001)."

The null and alternative hypothesis are:

$H_0: \mu_r-\mu_b=0\\\\H_1: \mu_r-\mu_b\neq0$

The level of significance is assumed to be α=0.01.

The difference of the sample means is:

$M_d=M_r-M_b=9.64-15.84=-6.20$

The standard deviation for the difference of the means is:

$s_d=\sqrt\frac{s_r^2+s_b^2}{n} } =\sqrt\frac{3.43^2+8.65^2}{21} }=\sqrt{\frac{85.59}{21} }=\sqrt{4.12} =2.03$

Then, the t-statistic is:

$t=\frac{M_d-(\mu_r-\mu_b)}{s_d} =\frac{-6.20-0}{2.03}= -3.05$

The degrees of freedom are:

$df=n_r+n_b-2=21+21-2=40$

With 40 degrees of freedom the t-critial for a significance of α=0.01 is t=±2.705.

As the t-statistic lies in the rejection region, the effect is significant and the null hypothesis is rejected.

There is enough statistical evidence to say that the means are different.

7 0

4 years ago

a tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an inital speed of 120 feet per second

balu736 [363]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{\textit{initial velocity of the object}}{120}\\\\
h_o=\stackrel{\textit{initial height of the object}}{2}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+120t+2$

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{llll}
h(t)= &{{ -16}}t^2&{{ +120}}t&{{ +2}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
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