Find the volume of this prism

The flagpole is 20 ft high and casts an 8 foot shadow, the oak tree casts a 12 foot shadow. How tall is the oak tree

Fudgin [204]

Answer: 4.8 ft

To answer this question you need to know how much shadow: actual height ratio. Flagpole is having 8 ft shadow with 20 ft actual height. The ratio should be= 8 ft: 20 ft= 0.4

Then multiply the ratio with the oak tree shadow. The equation would be:
0.4 x 12 ft= 4.8 ft

7 0

4 years ago

Find the lateral surface area of the prism below in square yards.

denis-greek [22]

Answer:

Step-by-step explanation:

Answer: 14*2 * 3

=> 6 * 14

<h2>=> 84</h2><h2></h2><h2>Please mark as brainliest for further answers :)</h2>

7 0

3 years ago

A 6 sided die is roller 4 times in a row. The probability of getting a 4 only on the last trial

yanalaym [24]

The answer would be 1 of 6

3 0

3 years ago

Write and solve a real - word problem that can be solved using the expression 3/4 divide by 1/6.

shepuryov [24]

A real word problem could be that max has 3/4 of a chocolate bar and he wants to divide the chocolate with 1/6 of her class. If he does that, how much will each person get?

4 0

4 years ago

A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio

yaroslaw [1]

Answer:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

$\hat p=0.45$ estimated proportion of chips that fail in the first 1000 hours of their use

$\mu_0 =0.48$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 claim that the true proportion si less then 0.48:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

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 is significantly different from a hypothesized value .

Calculate the statistic

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

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

6 0

4 years ago