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3 years ago

Rectangle A has length 12 and width 8. Rectangle B has length 15 and width 10. Rectangle C has length 30 and width 15.

Mathematics

1 answer:

Anna007 [38]3 years ago

7 0

Answer:

yes; 1.25

Step-by-step explanation:

The length to width ratios of the rectangles are ...

A: 12/8 = 1.5

B: 15/10 = 1.5

C: 30/15 = 2.0

Rectangles A and B have the same aspect ratio, so are similar. Rectangle B is a scaled copy of A with a scale factor of 10/8 = 1.25.

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NEED HELP ASAP PLEASE

Ilya [14]

Answer:

see explanation

Step-by-step explanation:

Given

A = $\frac{1}{2}$ bh

a. substitute b = 6 and h = 5 into the formula

A = $\frac{1}{2}$ × 6 × 5 = $\frac{1}{2}$ × 30 = 15 units²

Multiply both sides of the formula by 2

2A = bh, that is

bh = 2A ← divide both sides by b

h = $\frac{2A}{b}$ = $\frac{2(16)}{4}$ = $\frac{32}{4}$ = 8

Using

bh = 2A ← divide both sides by h

b = $\frac{2A}{h}$ = $\frac{2(12)}{4}$ = $\frac{24}{4}$ = 6

3 0

3 years ago

For the hypothesis test H0: μ = 10 against H1: μ >10 and variance known, calculate the Pvalue for each of the following test

Basile [38]

Answer:

a) $p_v =P(Z>2.05)=1-P(z$

b) $p_v =P(Z>-1.84)=1-P(z$

c) $p_v =P(Z>0.4)=1-P(z$

Step-by-step explanation:

Some previous concepts

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct.

A z-test for one mean "is a hypothesis test that attempts to make a claim about the population mean(μ)".

The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"

The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"

Hypothesis

Null hypothesis: $\mu=10$

Alternative hypothesis: $\mu >10$

If the random variable is distributed like this: $X \sim N(\mu,\sigma)$

We assume that the variance is known so the correct test to apply here is the z test to compare means, the statistic is given by the following formula:

$z_o=\frac{\bar X -\mu}{\sigma}$

Since we have the values for the statistic already calculated we can calculate the p value using the following formulas:

Part a

$p_v =P(Z>2.05)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(2.05,0,1,TRUE)"

Part b

$p_v =P(Z>-1.84)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(-1.84,0,1,TRUE)"

Part c

$p_v =P(Z>0.4)=1-P(z$

And in order to find the answer using excel we can use the following code:

"=1-NORM.DIST(0.4,0,1,TRUE)"

Conclusions

If we use a reference value for the significance, let's say $\alpha=0.05$ . For part a the $p_v$ so then we can reject the null hypothesis at this significance level.