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3 years ago

The relationship between the number of $4 lunches you buy with $100 lunch card and the money remaining on the cards.

Mathematics

1 answer:

yanalaym [24]3 years ago

5 0

$4x = $100

100 /4 is 25. You can by 25 lunches....

hope you have a few friends to share it with lol.

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When we reject the null and the null is true, we have a made a _________ __________.

nydimaria [60]

When we reject the null and the null is true, we have a made a type I error

The null hypothesis in statistics states that there is no difference between groups or no relationship between variables. It is one of two mutually exclusive hypotheses about a population in a hypothesis test.

null hypothesis is denoted as H₀

Reject the null hypothesis when the p-value is less than or equal to your significance level. Your sample data favor the alternative hypothesis, which suggests that the effect exists in the population. When you can reject the null hypothesis, your results are statistically significant.

when the p-value is greater than your significance level, you fail to reject the null hypothesis.

Sometimes , we reject our null hypothesis even when its true

there we made a type I error in hypothesis

To know more about null hypothesis here

brainly.com/question/19263925

#SPJ4

3 0

1 year ago

Plz help , due in 5 mins

Grace [21]

Answer:

option a) 30 : 19

Step-by-step explanation:

$Length = 7 \frac{1}{2} = \frac{15}{2}\\\\Width = 4\frac{3}{4} = \frac{19}{4}$

Ratio of length to width:

$\frac{15}{2} : \frac{19}{4}$

$=\frac{\frac{15}{2}}{\frac{19}{4}}\\\\=\frac{15}{2 } \times \frac{4}{19}\\\\=15\times \frac{2}{19}\\\\=\frac{30}{19}\\\\30 : 19$

3 0

2 years ago

Complete the sentence.

Over [174]

Answer:

Step-by-step explanation:

The horizontal asymtote is when there are no solutions for x, so it is when y = -10.

This is because when y = -10, 0.8^x must be 0, but this is impossible. You can test this by trying 4^x/5^x. It will get super small, but never 0.

5 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago

tony and some friends went to the movies. They bought 4 drinks and 2 tubs of popcorn and spent a total of $32.50 on the food. Ea

poizon [28]

X=popcorn
x - 3.50=drink
2x+ 4(x-3.50)=32.50
2x+4x-14=32.50

solve for x

8 0

3 years ago

Read 2 more answers