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Aliun [14]
1 year ago
12

H(x) = -2x+5 g(x) = 2x² + x Find (h+g)(1)

Mathematics
1 answer:
Westkost [7]1 year ago
4 0

Are you evaluating functions for this? If so, here you go:

Given the functions to simplify (h + g)(1), the answer you seek would be 6 I believe.

Hope this helps!

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Triss [41]
9-5=4 Therefore f=4
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Which Expression is the best estimate of 1 1/5 x 3 2/3?
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The best estimate would be 121/15 = 8.1
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- 4 ( x - 5 ) = 92<br><br> What is x?
Stells [14]

Answer:

Let's solve your equation step-by-step.

−4(x−5)=92

Answer:

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Step-by-step explanation:

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3 years ago
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A parachute speed during a free fall reaches117 miles per hour What is the speed in feet per second At the speed how many feet w
qwelly [4]

Answer:

Speed is 171.6 ft/s and distance is 1716 feet

Step-by-step explanation:

Given that,

The speed of a parachute = 117 miles per hour.

1 mph = 1.47 ft/s

117 mph = 171.6 ft/s

We need to find the distance it travel during 10 s. We know that the distance covered is equal to the product of speed and time taken. So,

d = vt

d=171.6\ ft/s\times 10\ s\\\\=1716\ \text{feet}

Hence, this is the required solution.

3 0
2 years ago
.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and
Leni [432]

Answer:

a) Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

b) t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

Step-by-step explanation:

Information given

\bar X=130 represent the sample mean for the amount spent each shopper

s=40 represent the sample standard deviation

n=80 sample size  

\mu_o =120 represent the value to verify

t would represent the statistic    

p_v represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

The statistic for this case would be given by:

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

Replacing the info given we got:

t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236  

The degrees of freedom are given by:

df = n-1 = 80-1=79

The p value for this case taking in count the alternative hypothesis would be:

p_v =P(t_{79}>2.236)=0.0141  

5 0
3 years ago
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