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3 years ago

Select the correct answer.

Mathematics

1 answer:

schepotkina [342]3 years ago

3 0

Answer:

He drove the rest of the trip at 30 miles per hour

Step-by-step explanation:

The rule of the distance is D = v × t, where

v is the speed

t is the time

∵ It is 170 miles from Bruce's house to the city where his brother lives

∴ D = 170 miles

∵ He drove for 2 hours at 55 miles per hour

∴ t1 = 2 hours and v1 = 55 miles/hour

→ By using the rule above find the distance of this part of his trip

∴ D1 = 55 × 2 = 110 miles

∵ The rest of the trip took 2 hours

∴ t2 = 2 hours

∵ D = D1 + D2

→ Substitute the values of D and D1 to find D2

∴ 170 = 110 + D2

→ Subtract 110 from both sides to find D2

∴ 60 = D2

∴ D2 = 170 - 110 = 60 miles

∵ The rest of the trip is 60 miles

∵ It took 2 hours

∵ D2 = v2 × t2

∴ 60 = v2 × 2

→ Divide both sides by 2

∴ 30 = v2

∴ v2 = 30 miles/hour

∴ He drove the rest of the trip at 30 miles per hour

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What is x^2 -7x + 12 = 0 what is using method of factoring

statuscvo [17]

$x^2 -7x + 12 = 0 \\ \\ (x - 4)(x - 3) = 0 \\ \\ x = 4, 3 \\ \\$

The final result is: x = 4, 3.

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3 years ago

Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in

Alborosie

Answer:

a) Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

b) $z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

$p_v =P(Z$

c) So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=0.46$ estimated proportion of American families owning stocks or stock funds

$p_o=0.53$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:

Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

Part b

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:

$z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

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 significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z$

Part c

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

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3 years ago

Add.<br><br> Express your answer in simplest form.<br> 2/9 + 4/9

timama [110]

2/9+4/9=6/9=2/3

Your answer is 2/3.

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3 years ago

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Katen [24]

Answer: