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xenn [34]
3 years ago
10

Lakita is going to plant trees and bushes to landscape her new yard. She cannot exceed her budget of $150. Each tree costs $45,

and each bush costs $20.
Which inequality correctly represents the situation, where t is the number of trees and b is the number of bushes Lakita purchases?

45t+20b>150

45t+20b≥150

45t+20b<150

45t+20b≤150
Mathematics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

C.45t + 20b ≤ 150

Step-by-step explanation:

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Need help with Algebra 2! Thank you! :)
Olin [163]

Answer:

Consider the parent logarithm function f(x) = log(x)

Now,

Let us make transformations in the function f(x) to get the function g(x)

•On streching the graph of f(x) = log(x) , vertically by a factor of 3, the graph of y = 3log(x) is obtained.

•Now, shrinking the graph of y = 3log(x) horizontally by a fctor of 2 to get the grpah of y = 3log(x/2) i.e the graph of g(x)

Hence, the function g(x) after the parent function f(x) = log(x) undergoes a vertical stretch by a factor of 3, and a horizontal shrink by a factor of 2 is

g(x) = 3 log(x/2) (Option-B).

7 0
2 years ago
A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. ​a) E
polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) \alpha =1-0.98=0.02

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq  p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and z_{\alpha/2} is the z-value that let a probability of \alpha/2 on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, \alpha by 0.02 and z_{\alpha/2} by 2.33

Where p' and \alpha are calculated as:

p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02

So, replacing the values we get:

0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq  p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence 1-\alpha is equal to 98%

It means the the level of significance \alpha is:

\alpha =1-0.98=0.02

4 0
3 years ago
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BigorU [14]

Answer:

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Step-by-step explanation:

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2 years ago
A microwave oven costs $124.85 before it is marked down. If it is marked down by 63%, what is its new price, to the
Aleksandr [31]

Answer:

D ($46.19)

Step-by-step explanation:

$124.85 * (100% - 63%)

= $124.85 * 37 %

= $124.85 * 0.37

= 46.1945

= $46.19 (to the nearest cent)

Hope this helped!

5 0
2 years ago
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