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dangina [55]
2 years ago
5

Ryan conducted a 6-day study observing the effects of an organic plant food on the growth of his sprouting bean plant. He tracke

d these two pieces of information:
the amount of plant food remaining in the container after each days feeding
the height of the plant over time
Ryan found that the amount of plant food remaining decreased an equal amount of each day, and he used the entire 72 milliliters by the end oh his study.
Which statement is true about the relationship between the amount of plant food remaining and the number of days?

This relationship is not a function because only one amount of plant food remains each day

This relationship is not a function because more that one amount of plant food remains each day

The relationship is a function because more than one amount of plant food remains each day

The relationship is a function because only one amount of plant food remains each day
Mathematics
1 answer:
LekaFEV [45]2 years ago
8 0

Answer:

The relationship is a function because only one amount of plant food remains each day

Step-by-step explanation:

The relationship is a function because only one amount of plant food remains each day

y (remaining at finish of the day) = - 12 x + 72

x = 1     y = 60

x = 2     y = 48

x = 3     y = 36

x = 4     y = 24

x = 5     y = 12

x = 6     y = 0

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Traffic speed: The mean speed for a sample of cars at a certain intersection was kilometers per hour with a standard deviation o
aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>

The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2} * Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }

t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71

[(33.99-26.56) ± 1.654 *(3.71*\sqrt{\frac{1}{135} +\frac{1}{42} })]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06

[(475.12-321.34) ± 2.485 *(33.06*\sqrt{\frac{1}{12} +\frac{1}{15} })]

[121.96; 185.60]

I hope this helps!

3 0
2 years ago
Find the 14th term of the geometric sequence 7, 28, 112, ...
bixtya [17]

Answer: five people like to like 5 people

Step-by-step explanation:

7 0
2 years ago
The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Po
Leona [35]

Answer:

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

Step-by-step explanation:

Data given

Campus   Sample size     Mean    Population deviation

   1                 330               33                      8

   2                310                31                       7

\bar X_{1}=33 represent the mean for sample 1  

\bar X_{2}=31 represent the mean for sample 2  

\sigma_{1}=8 represent the population standard deviation for 1  

\sigma_{2}=7 represent the population standard deviation for 2  

n_{1}=330 sample size for the group 1  

n_{2}=310 sample size for the group 2  

\alpha Significance level provided  

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37  

P value  

Since is a one right tailed test the p value would be:  

p_v =P(Z>3.37)=1-P(Z  

Comparing the p value with a significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.  

5 0
2 years ago
5. What is the area of the figure to the nearest square centimeter? It is composed of a symmetric hexagonand a semicircle 4 cm6
Nookie1986 [14]
=100.2743338823 which is 100
6 0
2 years ago
The fish population of Lake Collins is decreasing at a rate of 3% per year. In 2004 there were about 1,300 fish. Write an expone
Sergeu [11.5K]
For this case we have a function of the form:
 y = A * (b) ^ x&#10; Where,
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 y = 1300 * (0.97) ^ x&#10;
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y = 1300 * (0.97) ^ x
 The population in 2010 is:
 y = 1083
6 0
2 years ago
Read 2 more answers
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