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2 years ago

Can you help me with this one

Mathematics

2 answers:

Vsevolod [243]2 years ago

6 0

Answer:

For number 2 It would be 10x+12=29

For number 3 Try -37 Because i'm ending up with -36

Step-by-step explanation:

If the 2 was outside of the parenthesis you would multiply both numbers in the parenthesis

Kitty [74]2 years ago

4 0

It c and b it won't let me write a lot :(

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Alicia drew a map of Texas. In the scale Alicia used for the map, .50 inch represents 20 miles. What is the distance between two

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Answer:

An actual distance of 360 miles correspond to 9 inches on the map

Step-by-step explanation:

From the question, we have a scale of 0.5 inches representing 20 miles

Now, we want to relate an actual distance of 360 miles to a measurement on the map using the scale

From here, we have that;

0.5 inches = 20 miles

x inches = 360 miles

x * 20 = 0.5 * 360

20x = 180

x = 180/20

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valentina_108 [34]

Answer:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

Step-by-step explanation:

Information given

60, 56, 60, 55, 70, 55, 60, and 55.

We can calculate the mean and deviation with these formulas:

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

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

Replacing we got:

$\bar X=58.875$ represent the mean

$s=5.083$ represent the sample standard deviation for the sample

$n=8$ sample size

$\mu_o =60$ represent the value that we want to test

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is less than 60, the system of hypothesis would be:

Null hypothesis: $\mu \geq 60$

Alternative hypothesis: $\mu < 60$

The statistic would be given by:

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

Replacing the info we got:

$t=\frac{58.875-60}{\frac{5.083}{\sqrt{8}}}=-0.626$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value would be given by:

$p_v =P(t_{(7)}$

Since the p value is higher than 0.1 we have enough evidence to FAIl to reject the null hypothesis and we can't conclude that the true mean is less than 60

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