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2 years ago

Carlos has money for a local hardware store. He uses his money to rent a lawn mower for 2 days. It costs

Mathematics

1 answer:

MrRa [10]2 years ago

7 0

Answer:

hi is this the answer?

Step-by-step explanation:

Over time, lawn mowers, pets and pick-up football games compact the ... They're available at rental centers, plus some hardware stores and garden centers. ... of the day, and a shaded lawn that may get only two to four hours of direct sun. ... If the water has not penetrated far enough, restart the watering and

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Which value of gg makes 26=7(g-9)+ 12 a true statement?

Nezavi [6.7K]

G=11 you add all the numbers together to get the correct number

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3 years ago

An apple orchard sells apples in bags of 10.the orchard sold a total of 2,430 apples one day.how many bags of apples waa

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243 bags of apples were sold in one day

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3 years ago

Help??

Diano4ka-milaya [45]

The tree is 52.8 feet, since the hypotenuse is only part of the tree, you have to add 12, since then it would equal the total height of the tree.

a^2+b^2=c^2
12^2+ 39^2=c^2
144+1521= 1665
√1665
= 40.8

40.8 + 12
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3 years ago

Read 2 more answers

Anyone know how to solve this?

Elina [12.6K]

35% of 80 is 28.

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I hope this helps.

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3 years ago

A recipe submitted to a magazine by one of its subscribers’ states that the mean baking time for a cheesecake is 55 minutes. A t

EastWind [94]

Answer:

$t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =P(t_{(9)}>4.296)=0.001$

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55.

Step-by-step explanation:

Information given

We have the following data: 54 55 58 59 59 60 61 61 62 65

The sample mean and deviation can be calculated with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X-i -\bar x)^2}{n-1}}$

$\bar X=59.4$ represent the sample mean

$s=3.239$ represent the sample standard deviation

$n=10$ sample size

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to test if the true mean is higher than 55, the system of hypothesis would be:

Null hypothesis: $\mu \leq 55$

Alternative hypothesis: $\mu > 55$

Replacing the info given we got:

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

And replacing the info given we got:

$t=\frac{59.4-55}{\frac{3.239}{\sqrt{10}}}=4.296$

The degrees of freedom are given by:

$df=n-1=10-1=9$

The p value for this case is given by:

$p_v =P(t_{(9)}>4.296)=0.001$

And for this case the p value is lower than the significance level so we have enough evidence to reject the null hypothesis and then we can conclude that true mean is higher than 55

8 0

3 years ago