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2 years ago

15

there are 55 students in a band 11 are 6th graders 21 are 7th grader and 23 are 8th graders what is the ratio of the 6th graders

in the band to the number of 7th and 8th graders in the band

1 answer:

Sonbull [250]2 years ago

8 0

Answer:

11 : 44

Step-by-step explanation:

There are 11 6th graders and a combination of 44 7th and 8th graders. Therefore, the ratio is 11 : 44. Hope this helps!

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3 years ago

Which function has a greater y-intercept?

Natali5045456 [20]

Answer:

Option C, both functions have an y-intersect equal to 2.

Step-by-step explanation:

When we have a function f(x), the y-intercept is the value f(0). This is the point where the graph of the function intersects the y-axis.

Then, for f(x) = -x^2 + 5*x + 2

The y-intercept is:

f(0) = -0^2 + 5*0 + 2 = 2

f(0) = 2

And for g(x) we do not have the equation, but we have the graph, so we can just look at which value of y does the graph intersects the y-axis.

We can see that the graph intersects the graph at y = 2

Then the y-intersect is: g(0) = 2

So both functions have the same y-intersect. Then the correct option is C.

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2 years ago

At the beginning of the week, Paloma opened a new box of 250 coffee creamers. Forty-four coffee creamers were used during the we

LiRa [457]

Hello, the correct answer would be B. Because if she had 250 then used 44 you would subtract 44 from 250 to find x

Hope this helps :)

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2 years ago

Read 2 more answers

The time it takes to travel a fixed distance varies inversely with the speed traveled. It takes

uysha [10]

first off, let's notice that Purple's time is in minutes, whilst the rate is in miles per hour, the units of both must correspond, so, we can either change the time from minutes to hours or the rate from hours to minutes, hmmm let's change the time to hours.

so 40 minutes, we know there are 60 minutes in 1 hour, so 40 minutes will be 40/60 of an hr, or namely 2/3.

$\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill$

$\stackrel{\begin{array}{llll} \textit{\tiny "t"ime varies}\\ \textit{\tiny inversely with "s"peed} \end{array}}{t = \cfrac{k}{s}}\qquad \textit{we know that} \begin{cases} t=\stackrel{minutes}{40}\to \stackrel{hrs}{\frac{2}{3}}\\ s=\stackrel{m/h}{9} \end{cases} \implies \cfrac{2}{3}~~ = ~~\cfrac{k}{9} \\\\\\ 18=3k\implies \cfrac{18}{3}=k\implies 6=k~\hfill \boxed{t=\cfrac{6}{s}} \\\\\\ \textit{when s = }\stackrel{m/h}{12}\textit{ what is "t"?}\qquad t=\cfrac{6}{12}\implies t=\cfrac{1}{2}$

6 0

2 years ago

2. A particular resource class must perform at a level consistent with the rest of the school to avoid being discontinued. On th

Natali5045456 [20]

Answer:

$t=\frac{71.86-78}{\frac{17.52}{\sqrt{14}}}=-1.311$

$p_v =P(t_{13}$

If we compare the p value and a significance level assumed for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the scores is NOT significant less than 78.

Step-by-step explanation:

Data given and notation

$\bar X=71.86$ represent the sample mean

$s=17.52$ represent the standard deviation for the sample

$n=14$ sample size

$\mu_o =78$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is less than 78, the system of hypothesis would be:

Null hypothesis: $\mu \geq 78$

Alternative hypothesis: $\mu < 78$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{71.86-78}{\frac{17.52}{\sqrt{14}}}=-1.311$

Calculate the P-value

First we need to find the degrees of freedom given by:

$df=n-1=14-1=13$

Since is a one-side left tailed test the p value would be:

$p_v =P(t_{13}$

Conclusion

If we compare the p value and a significance level assumed for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the scores is NOT significant less than 78.

3 0

3 years ago