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AnnZ [28]
3 years ago
7

Bill spent less than 26$ on a magazine and five books. The magazine cost 4$

Mathematics
2 answers:
mamaluj [8]3 years ago
5 0

Answer:

24 dollars

Step-by-step explanation:

6$for each one if u multiply then you'll get the number

bija089 [108]3 years ago
5 0

Answer:

$24.00

Step-by-step explanation:

hope i helped

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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Ipatiy [6.2K]

Answer:

x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}

Step-by-step explanation:

2x^2 - 10x - 3 = 0 \\\\a = 2 \ , b = - 10 \ , \ c =  - 3 \\\\x = \frac{-b^2\  \pm \ \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{10 \ \pm \sqrt{(-10)^2 - ( 4 \times 2 \times -3)} }{2 \times 2}\\\\x = \frac{10 \ \pm \sqrt{(100 - ( -24 )} }{4}\\\\x = \frac{10 \ \pm \sqrt{(100 + 24 } }{4}\\\\x = \frac{ 10 \ \pm \sqrt{124}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{4 \times 31}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{2^2 \times 31}}{4}\\\\x = \frac{ 10 \ \pm2 \sqrt{31}}{4}\\\\x = \frac{ 5 \ \pm\sqrt{31}}{2}\\\\

x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}

3 0
3 years ago
A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
F(x)=2x+3/4x+5<br>find f(-9)<br>​
mylen [45]

\implies {\blue {\boxed {\boxed {\purple {\sf { f(-9)= 0.48}}}}}}

\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}

f(x) =  \frac{2x + 3}{4x + 5} \\

For f(-9), put "-9" for every value of "x".

↬f( - 9) =  \frac{2( - 9) + 3}{4( - 9) + 5}\\

↬ f(-9) =  \frac{ - 18 + 3}{ - 36 + 5} \\

↬ f(-9) =  \frac{ - 15}{ - 31}\\

↬ f(-9)=  \frac{15}{31}\\

↬f(-9)=  0.48\\

\bold{ \green{ \star{ \red{Mystique35}}}}⋆

6 0
3 years ago
Read 2 more answers
Please help im doing a test and it’s due soon
swat32

Answer:

the answer is D.

Step-by-step explanation:

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7 0
3 years ago
A video has 42 thumbs down. What integer represents its score in points?
ioda

Answer:

Step-by-step explanation: it is negative 42 it is just the opposite cause the thumbs is down negative up positive

6 0
3 years ago
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