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3 years ago

If g(x)=5x-3 and h(x)=square roto of x, find (g•h) (4)

Mathematics

1 answer:

denis-greek [22]3 years ago

3 0

Answer:

(goh) (4) = 7

Step-by-step explanation:

(goh) (x) = g(h(x))

Plug in

g(h(x)) = 5(√x) - 3

if x = 4 then

g(h(4)) = 5(√4) - 3

g(h(4)) = 5(2) - 3

g(h(4)) = 10 - 3

g(h(4)) = 7

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Expand the expression 5(p - 1) + 3(p - 0.75)

Whitepunk [10]

Answer:

=8p−7.25

Step-by-step explanation:

=(5)(p)+(5)(−1)+(3)(p)+(3)(−0.75)

=5p+−5+3p+−2.25

=(5p+3p)+(−5+−2.25)

=8p+−7.25

6 0

3 years ago

If clocks read 300pm on the -5 time zone, what time do the clocks read in the -2 time zone

gayaneshka [121]

It will read 6:00pm on -2 time zone.

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3 years ago

The graphs below have the same shape. The equation of the blue graph is f(x)

Ghella [55]

Answer:

Step-by-step explanation:

I attached the graphs

7 0

2 years ago

Find the measures of interior angles. Find a and B.

olga nikolaevna [1]

Answer:

A = 68

B = 44

C = 68

Step-by-step explanation:

find x:

68 + x + x - 24 = 180

2x + 44 = 180

180 - 44 = 136

136 / 2 = 68

x = 68

_________________________

68 + 68 + 44 = 180

6 0

3 years ago

A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz

marin [14]

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean, $\mu_x$ is calculated as:

$\mu_x=\mu=204.9, \mu_x=sample \ mean$

-The standard deviation, $\sigma_x$ is calculated as:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968$

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean, $\mu_x$ is calculated as:

$\mu_x=\mu\\\\=75.9$

#The sample standard deviation is calculated as follows:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136$

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

$\mu_x=\mu=135.7$

#Sample standard deviation:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566$

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

$P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452$

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

7 0

3 years ago