Ask question

Ask question

All categories

3 years ago

14

Marcia has a sunflower garden. The plants are currently 36 inches tall and are growing at a rate of 4 inches each week. Write an

equation that represents the height, H, of the sunflower plants, where w is the number of weeks.

1 answer:

alexgriva [62]3 years ago

6 0

H = 4w + 36
is your equation

You might be interested in

Which choice is equivalent to 6.25% ?

Levart [38]

.625 to find the decimal of a percentage divide by 100

6 0

3 years ago

An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry ou

Katyanochek1 [597]

Answer:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

$p_v =P(t_{(5)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation

The mean and sample deviation can be calculated from the following formulas:

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

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

$\bar X=29.35$ represent the sample mean

$s=1.365$ represent the sample standard deviation

$n=6$ sample size

$\mu_o =30$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :

Null hypothesis: $\mu \geq 30$

Alternative hypothesis: $\mu < 30$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=6-1=5$

Since is a one-side left tailed test the p value would given by:

$p_v =P(t_{(5)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

5 0

3 years ago

Find the area of the figure below

Serhud [2]

Easy

the easiest way is to imagine that you have a rectangle with a corner cut off
the area of the figure=area of rectangle-area cut off

area that was cut off was a triangle

area of rectangle=length times width
area of triangle=1/2base times height

so we draw imaginary lines like in attachment
find area of rectangle
legnth=9 inch
width=9 inch
area=9 times 9=81 in^2

triangle
the top part is ?+5=9, so the base of triangle=4
the side is ?+4=9, so the base is 5
area=1/2 times 4 times 5=10 in^2

area=rectangle-triangle
area=81-10=71 in^2

7 0

3 years ago

Read 2 more answers

Write the equation of line whose b= -1 and m=5

vovangra [49]

Answer:

Step-by-step explanation:

y=mx+b

y=5x-1

4 0

3 years ago

Read 2 more answers

Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (

Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

$P(\frac{T}{S}) = 0.95$

$P(\frac{T}{S^c}) = 0.05$

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

$P(S^c) = 1 - P(S)$

= 1 - 0.3

= 0.7

ii. $P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}$

$= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }$

$= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}$

= 0.8906

iii. $P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}$

$= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }$

$= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}$

= 0.0221

We simply applied the above formulas so that the each part could come

8 0

3 years ago