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larisa86 [58]
3 years ago
6

Eva is going to the amusement park, where she has to pay a set price of admission and another price for tickets to go on each of

the rides. The total amount of money Eva will spend is given by the equation y=26+1.5x. , where y represents the total amount of money, in dollars and cents, and represents the number of rides Eva goes on . What could the number 1.5 represent in the equation ?

Mathematics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

B. The change in the total amount if money for every one additional ride she goes on

Step-by-step explanation:

Given,

y = 26 + 1.5x, where,

y = total amount of money Eva will spend

x = number of rides Eva rides on

Thus,

Set price of admission into the amusement park = $26

Price per ride = $1.5

Therefore, the 1.5 I'm the equation can be said to be "the change in the total amount if money for every one additional ride she goes on".

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To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo
mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}

Given:

\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25

Confidence level = 80%

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0
3 years ago
An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
Veseljchak [2.6K]

Answer:

<em>The calculated value |t| = 2.375 > 2.1009  at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>Alternative hypothesis is accepted</em>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em /></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given data</em>

<em>First sample size (n₁) = 10 </em>

<em>mean of the first sample(x₁⁻) = 537</em>

<em>standard deviation of the first sample (S₁) = 22</em>

<em>second sample size (n₂) = 10 </em>

<em>mean of the second sample (x₂⁻) = 559</em>

<em>standard deviation of the second sample (S₂) = 17</em>

<u><em>Step(ii)</em></u><em>:-</em>

<u><em>Null hypothesis : H₀:</em></u><em>- The extra carbonation of cola results in a lower average compression strength</em>

<u>Alternative Hypothesis :H₁</u>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em>Step(iii)</em></u><em>:-</em>

<em>By using student's t -test for difference of means</em>

<u><em>Test statistic</em></u>

<em>       </em>t = \frac{x^{-} _{1}-x^{-} _{2}  }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} }  } )}<em />

<em>  where </em>

<em>     </em>S^{2}  = \frac{n_{1}S^{2} _{1} + n_{2} S_{2} ^{2}  }{n_{1}+n_{2} -2 }<em />

<em>    </em>S^{2}  = \frac{10(22)^{2}  + 10 (17) ^{2}  }{10+10 -2 } = \frac{ 7730}{18} = 429.4<em />

<em>    </em>t = \frac{537-559 }{\sqrt{429.4 (\frac{1}{10 }+\frac{1}{10 }  } )}<em />

<em>   t =  -2.375</em>

<em>|t| = |-2.375| = 2.375</em>

<em>Degrees of freedom</em>

<em>γ = n₁+n₂ -2 = 10+10-2 =18</em>

<em />t_{\frac{\alpha }{2} ,n-1}=t_{(\frac{0.05}{2} ,18)} = t_{(0.025,18)}}=2.1009<em />

<em>The calculated value |t| = 2.375 > 2.1009  at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>Alternative hypothesis is accepted</em>

<u><em>Final answer:-</em></u>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em /></u>

<em />

<em> </em>

<em />

5 0
3 years ago
HELP PLEASE!!!
ch4aika [34]

Answer:

The population of Bear in 2050 is 4750000

Step-by-step explanation:

A) The exponential growth equation for bear is as follows -

dN/dT = rmax * N

Where dN/dT = change in population

rmax is the maximum rate of change

N = Base population

B) Here the per capita rate of increase (r) will always be a positive value irrespective of the and hence we will assume this population to be growing exponentially.

C) dN/dT = rmax * N

D) dN / 5 = 2.5 * 380,000

dN = 5*2.5 * 380000

= 4750000

8 0
3 years ago
Dan earns £9.30 per hour. How much will he earn for 9 hours work?
Dimas [21]

Answer: £83.70

Step-by-step explanation:

We can set up a ratio.

Let x be the amount he earns for 9 hours of work.

\frac{9.30 pounds}{1 hour} = \frac{x pounds}{9 hours}

We can cross multiply:

x = 9*9.3

x = £83.70

6 0
2 years ago
Calculate the product 48.9.8
Brilliant_brown [7]
Do you want it to multiply. O que ?
3 0
4 years ago
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