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3 years ago

The graph shows the number of paintballs a machine launches, y, in x seconds:

Mathematics

1 answer:

aalyn [17]3 years ago

5 0

The answer is B, 20 balls in 4 seconds. Or 5 balls per second.

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A pizza place offers the following toppings: pepperoni, sausage, onions. green peppers, and mushrooms.

Harman [31]

Answer:

Step-by-step explanation:

You do 5C3, which is 5!/((5-3)!3!) -->

5*4*3*2*1/(2*1*3*2*1) =

5 0

3 years ago

Is there statistically significant evidence that the districts with smaller classes have higher average test scores? The t-sta

Musya8 [376]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval is $[670.03 , 673.97 ]$

The test statistics is $t = 7.7$

The p-value is $p-value = 0$

The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score

Step-by-step explanation:

From the question we are told that

The sample size is n = 408

The sample mean is $\= y = 672.0$

The standard deviation is $s = 20.3$

Given that the confidence level is 95% then the level of significance is

$\alpha = (100 -95 )\% = 0.05$

From the normal distribution table the critical value of $\frac{\alpha }{2} = \frac{0.05 }{2}$ is

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

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }$

=> $E = 1.96 * \frac{20.3}{\sqrt{408} }$

=> $E = 1.970$

Generally the 95% confidence interval is mathematically represented as

$\= y -E < \mu < \= y + E$

=> $672.0 -1.970 < \mu < 672.0 +1.970$

=> $670.03 < \mu < 673.97$

=> $[670.03 , 673.97 ]$

From the question we are told that

Class size small large

$Avg.score(\= y)$ $\= y_1 = 683.7$ $\= y_2 = 676.0$

$S_y$ $S_{y_1} =20.2$ $S_{y_2} = 18.6$

sample size $n_1 = 229$ $n_2 = 184$

The null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the standard error for the difference in mean is mathematically represented as

$SE = \sqrt{\frac{S_{y_1}^2 }{n_1} +\frac{S_{y_2}^2 }{n_2} }$

=> $SE = \sqrt{20.2^2 }{229} +\frac{18.6^2 }{184_2} }$

=> $SE = 1.913$

Generally the test statistics is mathematically represented as

$t = \frac{\= y _1 - \= y_2 }{SE}$

=> $t = \frac{683.7 - 676.0 }{1.913}$

=> $t = 7.7$

Generally the p-value is mathematically represented as

$p-value = P(t > 7.7 )$

From the z-table

$P(t > 7.7 ) = 0$

$p-value = 0$

From the values we obtained and calculated we can see that $p-value < \alpha$

This mean that

The p-value suggests that the null hypothesis is rejected with a high degree of confidence. Hence there is statistically significant evidence that the districts with smaller classes have higher average test score

4 0

3 years ago

Linear Functions! Step by step Solution for 4(x+2)-1/2x=22 equation Thankss❤️

Elena L [17]

4(x + 2) - 1/2x = 22
4x + 8 - 1/2x = 22
3.5x + 8 = 22
-8 -8
3.5x = 14
/ 3.5 /3.5
x = 4

6 0

3 years ago

Read 2 more answers

BRAINLIEST AND 40 POINTS PLZ HELP I NEVER GET HELP

mezya [45]

6.40 units

You use the squaring method to fill in the triangle.

Count the sides 4 up and 5 across
Square them (16 and 25)
Add them together 16+25=41
Find the square root of 41 (6.40)

Good luck!

6 0

2 years ago

Read 2 more answers

Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0

Xelga [282]

Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
$\left(h+c,\:k\right),\:\left(h-c,\:k\right)$

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}$
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

$9x^2-y^2-36x-4y+23=0 \ \textgreater \ \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}$
$9x^2-36x-4y-y^2=-23$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms}$
$9\left(x^2-4x\right)-\left(y^2+4y\right)=-23$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1$
$\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)$

$\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \ \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \ Refine$
$\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1$

Now rewrite in hyperbola standardform
$\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1$

$\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3$
$\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)$

Now we must compute c
$\sqrt{1^2+3^2} \ \textgreater \ \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \ 1^2 = 1 \ \textgreater \ \sqrt{1+3^2}$

$3^2 = 9 \ \textgreater \ \sqrt{1+9} \ \textgreater \ \sqrt{10}$

Therefore the hyperbola foci is at $\left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)$

For the vertices we have $\left(2+1,\:-2\right),\:\left(2-1,\:-2\right)$

Simply refine it
$\left(3,\:-2\right),\:\left(1,\:-2\right)$
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0

3 years ago