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1 year ago

Let f(x)=3/4x2-1 The function g(x) is a vertical stretch of f(x) by a factor of 8.What is the equation of g(x)?

Mathematics

1 answer:

mote1985 [20]1 year ago

3 0

Answer:

g(x) = 6x^2-1

Step-by-step explanation:

f(x)=3/4x^2 −1.

A vertical stretch by a factor of 8 you multiply the number with x by the factor:

3/4 x 8 = 6

g(x) = 6x^2-1

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b. Find a 99% confidence interval for mean age of ICU patients using the facts that, in the sample, the mean is 57.55 years and

rjkz [21]

Answer:

99% of confidence intervals for mean age of ICU patients

(53.8920 , 61.2079)

Step-by-step explanation:

Explanation:-

Given sample mean $x^{-} = 57.55$

Given standard error is determined by

S.E = $\frac{S.D}{\sqrt{n} }$

Given data standard error = 1.42

99% of confidence intervals for mean is determined by

$(x^{-} - Z_{0.01} \frac{S.D}{\sqrt{n} } , x^{-} + Z_{0.01} \frac{S.D}{\sqrt{n} } )$

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.05} = 2.576$

$(x^{-} - 2.576 S.E , x^{-} + 2.576 S.E)$

$(57.55 - 2.576X 1.42 , 57.55+ 2.576 X1.42)$

(53.8920 , 61.2079)

Conclusion:-

99% of confidence intervals for mean is determined by

(53.8920 , 61.2079)

5 0

2 years ago

You and a friend went to the movies.You spent half your money on tickets. Then you spent 1/4 of the money on popcorn, $2 on cand

bagirrra123 [75]

Answer:

We took $ 24 to the movies.

Step-by-step explanation:

Let total money be

On movies = Half of the money = $\frac{x}{2}$

On popcorn = $\frac{x}{4}$

on candy = $2

On soda = $3

total money spent = $\frac{x}{2}$ + $\frac{x}{4}$ + 2+3+1

x = $\frac{x}{2}$ + $\frac{x}{4}$ + 2+3+1

x- $\frac{x}{2}$ - $\frac{x}{4}$ = 6

$\frac{x}{1}$ - $\frac{x}{2}$ - $\frac{x}{4}$ =6

$\frac{4x-2x-x}{4} = 6\\ \frac{x}{4} =6\\x =24$

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2 years ago

The shirt cost $113 less than the coat. The shirt cost $50. How much is the coat?

olga nikolaevna [1]

Shirt: $50
Coat: $113

$50+ $113= $163

The coat costs $163, hope this helps!

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Two and six hundred thousandths in standard form

mihalych1998 [28]

2.600 in standard form

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