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3 years ago

Solve the following and

Mathematics

1 answer:

sladkih [1.3K]3 years ago

4 0

Answer:

7.5* 10^8

Step-by-step explanation:

(1.5 x 10^5)(5 x 10^3)

Multiply the numbers

1.5*5=7.5

Add the exponents

10 ^(5+3) = 10^8

Put back together

7.5* 10^8

This is in scientific notation

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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

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3 years ago