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2 years ago

1. f(x) = 5x 2. g(x) = 54 What are the ranges for these functions

Mathematics

1 answer:

STatiana [176]2 years ago

4 0

Answer:

For 1: y is all real numbers. For 2, y=54.

Step-by-step explanation:

1: Since the graph is linear and infinite, there will be at least 1 y value that corresponds with any x value.

2: Since range measures only the y values, and since g(x)=54 is a straight line, it only has a value of y=54 for any values of x.

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I have 30 coins consisting of nickels, dimes and quarters. The total value of the coins is $4.60. there are two more dimes than

aleksklad [387]

If you have 1 nickle, how many quarters do you have? (3)

If you have 4 nickles, you have 3 times as many quarters (3)(4) = 12

If you have n nickles, then you have 3n quarters so 3n = q, you have a slightly different equation.

If you fix this equation and use substitution like you did, you can get the right answer; you can also try to work in the other information that you have - converting all coin values to cents

5n + 10d + 25q = 460

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2 years ago

Solve: 3 ln x = ln 216

BaLLatris [955]

$3\ln x=\ln 216\\ \ln x^3 =\ln 216\\ x^3=216\\ x=6$

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3 years ago

Read 2 more answers

A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salari

guajiro [1.7K]

Answer:

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

$F_{critical} = 1.8608$

$F = 2.9085$

The decision rule is

Reject the null hypothesis

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 30$

The sample variance for elementary school is $s^2_1 = 8324$

The second sample size is $n_2 = 30$

The sample variance for the secondary school is $s^2_2 = 2862$

The significance level is $\alpha = 0.05$

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

Generally from the F statistics table the critical value of $\alpha = 0.05$ at first and second degree of freedom $df_1 = n_1 - 1 = 30 - 1 = 29$ and $df_2 = n_2 - 1 = 30 - 1 = 29$ is

$F_{critical} = 1.8608$

Generally the test statistics is mathematically represented as

$F = \frac{s_1^2 }{s_2^2}$

=> $F = \frac{8324 }{2862}$

=> $F = 2.9085$

Generally from the value obtained we see that $F > F_{critical }$ Hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is sufficient evidence to support the researchers claim

4 0

2 years ago

insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of

ss7ja [257]

Answer: $(13.22\%,\ 19.34\%)$

Step-by-step explanation:

Given : Sample space : n= 559

Sample proportion : $\hat{p}=\dfrac{91}{559}=0.162790697674\approx0.1628$

$=16.28\%$

Significance level : $\alpha= 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Confidence level for population proportion:-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.1628\pm (1.96)\sqrt{\dfrac{0.1628(1-0.1628)}{559}}\\\\=0.1628\pm0.030604971367\\\\\approx 0.1628\pm0.0306=(0.1322,\ 0.1934)=(13.22\%,\ 19.34\%)$

Hence, 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.= $(13.22\%,\ 19.34\%)$

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3 years ago

ADP and ABQ are tangents to the circle, centre O.

Anarel [89]

Answer:

Step-by-step explanation:

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2 years ago