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2 years ago

If f(x) = 5x + 40, what is f(x) when x = –5? a.–9 b.–8 c.7 d.15

Mathematics

2 answers:

Helen [10]2 years ago

5 0

Answer:

f(x)=15 d.

Step-by-step explanation:

To begin solving the equation you need to substitute -5 in for x.

The equation would then read f(x)=5(-5)+40

Then you need to simplify 5 times -5 is -25

the equation would be f(x)=-25+40

Then you would simplify and you will get f(x)=15

defon2 years ago

3 0

Just plug the number

_5(5)+40=-25+40

=15

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What is the constant of the polynomial 2x3 - 8x2 + 3x - 7

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Answer:

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Step-by-step explanation:

A constant number is a number that contains no variables like x and y. The only constant in that problem is -7.

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Each journal weighs about 280 grams.What is total weight of 3 jpurnals?

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1 year ago

A biologist is studying the growth of a particular species of algae. She writes the following equation to show the radius of the

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Let's solve for d.

fd=(7)(1.06)d

Step 1: Add -7.42d to both sides.

df+−7.42d=7.42d+−7.42d

df−7.42d=0

Step 2: Factor out variable d.

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Step 3: Divide both sides by f-7.42.

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1 year ago

A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

Citrus2011 [14]

Answer:

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b) 95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

98% of confidence intervals

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

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

$(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )$

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

95% of confidence intervals

$({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } } )$

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36 (check table)

X^2_{0.95,13} = 5.892 (check table)

$(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13} } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )$

$(297.29. (\sqrt{\frac{14-1}{22.36 } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )$

(214.91 , 441.53)

95% confidence interval for the standard deviation.

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