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3 years ago

What are the domain and range of the function f(x)=3/4x+5?

Mathematics

1 answer:

dybincka [34]3 years ago

7 0

Start with parentheses

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I need help on all of those. Pleeeaaassseee help I really need it

goldenfox [79]

39.7 times 100, or:

3970 inches squared

7 0

3 years ago

3⋅f(−4)−3⋅g(−2)= -27

Reil [10]

Answer:

3⋅f(−4)−3⋅g(−2)= -27 is g = -9/2 +2f

Graph the line using the slope and y-intercept, or two points.

Slope:

6 0

3 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

Given that the level of confidence is 90% the significance level is mathematically evaluated as

$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

4 0

3 years ago

Math help @Alessandroarenas

frez [133]

Accordinly to the details, i think that A could work.

7 0

3 years ago

Eric rented a truck for one day the base fee was 16.95 for additional charge of .76 cents for each mile driven eric had to pay 1

goldenfox [79]

Answer:

178 miles

Step-by-step explanation:

Let Eric has driven d miles.

As the rate of charge of driving the truck = 76 cents /mile

So, the charge of driving the truck for d miles = 76d cents= $0.76d

The base fee of the truck was $16.95.

Total money paid at the time of returning the truck = $152.23

So, 16.95+0.76d=152.23

0.76d= 152.23-16.95=135.28

d=135.28/0.76

d=178

Hence, he drives 178 miles.

8 0

2 years ago