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AlekseyPX
3 years ago
8

A school club sold 220 T-shirts for a fundraiser. Adult-size shirts cost $20 each and child-size shirts cost $15 each. The total

collected for all the T-shirts sold was $3,550.
How many adult-size T-shirts were sold?
Mathematics
2 answers:
andre [41]3 years ago
4 0
{20a+15c=3550
{a+c=220

{20a+15c=3550
{15a+15c=3300
5a=250
a=50
There are 50 adult-size T-shirts were sold
pishuonlain [190]3 years ago
3 0

Answer: There are 50 adult-size shirts that were sold .

Step-by-step explanation:

Since we have given that

Cost of adult-size shirts = $20

cost of child-size shirts =$15

Let hte number of adult-size shirt be x

Let the number of child-size shirt be y

According to question,

20x+15y=\$3550----------------------------(1)\\\\x+y=220---------------------------(2)\\\\

So, we will use "Substitution method" :

x+y=220\\x=220-y

Put this value in equation (1):

20x+15y=3550\\\\20(220-y)+15y=3550\\\\4400-20y+15y=3550\\\\4400-5y=3550\\\\4400-3550=5y\\\\850=5y\\\\\frac{850}{5}=y\\\\170=y

So, put the value of y in x :

x=220-y\\\\x=220-170\\\\x=50

Hence, there are 50 adult-size shirts that were sold .

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A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
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a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

<u>Step-by-step explanation</u>:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

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A survey of 25 young professionals statistically that the population mean is less than $30

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