the temperature in Portland, Marine, reached -3.5ºF one day last winter. Between with two integers does the temperature lie?

A student took a system of equations, multiplied the first equation by 2 and the second equation by 3, then added the results to

egoroff_w [7]

Answer:

system 4.

Step-by-step explanation:

3x+6y=9

multiply by 2

6x+12y=18 ....(1)

-2x+-4y=4

multiply by 3

-6x-12y=12 ...(2)

add (1) and (2)

0+0=30

which is impossible .It has no solutions.

3 0

2 years ago

Please help!!! will mark brainliest

faust18 [17]

Your answer is 7.

I subtracted 5 from 12 to find the missing number for KL that is 7.

Therefore, your answer is 7

6 0

1 year ago

Read 2 more answers

What fraction is equal to 4/6

Rama09 [41]

4/6
= (4/2) / (6/2) (divide both numerator and denominator by 2)
= 2/3

The final answer is 2/3~

6 0

2 years ago

Solve forn: -6(n-8) = 4(12-5n)+14n

Irina-Kira [14]

Answer:

0

Step-by-step explanation:

-6(n-8) = 4(12-5n)+14n

-6(n-8)=-6n+48 = 4(12-5n)+14n= 48-20n+14n

then 48-20n+14n<em> = 48-6n</em>

so you have

-6n+48 = <em>48-6n</em>

-6n+6n= n

n+48=48

n-48=48-48

n=0

5 0

3 years ago

Read 2 more answers

A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located

GenaCL600 [577]

Answer:

$z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589$

$p_v =P(z>1.589)=0.056$

If we compare the p value obtained with the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

Step-by-step explanation:

Data given and notation

n=497 represent the random sample taken

X=340 represent the homes in Omaha with one or more lawn mowers

$\hat p=\frac{340}{497}=0.684$ estimated proportion of homes in Omaha with one or more lawn mowers

$p_o=0.65$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:

Null hypothesis: $p\leq 0.65$

Alternative hypothesis: $p > 0.65$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.589)=0.056$

If we compare the p value obtained with the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

7 0

2 years ago