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sesenic [268]
2 years ago
12

This equation shows how the number of pies Larry can bake is related to the number of additional cups of sugar he buys.

Mathematics
1 answer:
Anna71 [15]2 years ago
7 0
Im pretty sure its 3
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Help me I really need help
natita [175]

Answer:

i think it would be the first one

Step-by-step explanation:

not a wild guess but i am definetly  not sure

8 0
3 years ago
Read 2 more answers
When your income is more than your expenses, you have _____.
blagie [28]
<span>When your income is more than your expenses, y</span>ou have surplus

3 0
3 years ago
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You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling
BARSIC [14]

Answer:

  (a) 15 m

  (b) ground speed: 26.14 m/s (94.1 km/h); relative speed 12 m/s

Step-by-step explanation:

We can choose the frame of reference to be the trailing car. We can start counting time from the point the lead car begins deceleration. Then its position (in meters) relative to the trailing car is ...

  d(t) = 25 - 5/2t² . . . . . . where 25 m is the initial distance and -5 is the acceleration in m/s².

(a) Then the distance between cars at t=2 is ...

  d(2) = 25 - (5/2)·4 = 15 . . . . meters

__

(b) The <em>relative velocity</em> at 2.4 seconds is ...

  d'(t) = -5t

  d'(2.4) = -5(2.4) = -12.0 . . . m/s

The closure speed with the police car is 12 m/s.

__

We need to do a little more work to find the ground speed of the trailing car when the cars bump.

The distance between the cars at t=2.4 seconds is ...

  d(2.4) = 25 -(5/2)2.4² = 10.6 . . . . meters

At the closure speed of 12 m/s, it will take an additional ...

  10.6 m/(12 m/s) = 0.8833... s = 53/60 s

to close the gap. The speed of the trailing car at that point in time will be the original speed less the deceleration for 0.8833 seconds:

  (110 km/h)·(1/3.6 (m/s)/(km/h))-(5 m/s²)(53/60 s)

     = 26 5/36 m/s = 94.1 km/h

_____

<em>Comment on following distance</em>

To avoid collision, the trailing car must be trailing by at least the distance it covers in 2.4 seconds, about 73 1/3 meters.

6 0
3 years ago
Log√30 - log√6 + log√2
Arada [10]

Assuming, it's decimal logarithm.

<h3>\log\sqrt{30}-\log\sqrt6+\log\sqrt2=\log\dfrac{\sqrt{30}\cdot\sqrt2}{\sqrt6}=\log \sqrt{10}=\dfrac{1}{2}\log 10=\dfrac{1}{2}\cdot 1=\dfrac{1}{2}</h3>
3 0
3 years ago
Suppose a poll is taken that shows that 765 out of 1500 randomly​ selected, independent people believe the rich should pay more
Zanzabum

Answer:

z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775  

p_v =P(z>0.775)=0.219  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.5

Step-by-step explanation:

Data given and notation

n=1500 represent the random sample taken

X=765 represent the successes

\hat p=\frac{765}{1500}=0.51 estimated proportion of successes

p_o=0.5 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 is higher than 0.5:  

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

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 \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

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

z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775  

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 right tailed test the p value would be:  

p_v =P(z>0.775)=0.219  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.5

8 0
3 years ago
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