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3 years ago

I need help with the range

Mathematics

1 answer:

seropon [69]3 years ago

8 0

Answer:

[-9, 0]

Step-by-step explanation:

The range is the y values that the equation/line covers.

The line's lowest point is -9 and the line's highest point is 0. The values are defined (no open circle at the highest point) so we use brackets.

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What is the average rate of change of f(x) over the interval from x=5 to x=9 is it -4 or 4?

ICE Princess25 [194]

In order to get the average rate you should do the y2-y1/x2-x1 formula. lets see our points are (5,-2)(9,14).. so our formula should be filled like 14+2/9-5, now subtract or add to get 16/4 now divide to get positive 4..

Hope this helps

7 0

3 years ago

Two times what wade's age was seven years ago

elena-s [515]

Answer:

how old is he now...?

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3 years ago

Identify the property that justifies each step asked about in the answer area below. Line 1: \text{Line 1: }\phantom{=}Line 1: =

kipiarov [429]

Answer:

1. Commutative Property of Addition

2. Distributive property

Step-by-step explanation:

Given

$Line 1: 4(8+9x)$

$Line 2: 4(9x+8)$

$Line 3: 36x+32$

Required

Justify each step

From Line 1 to 2:

The property from Line 1 to 2 depicts $a + b = b + a$ ;

i.e.

$4(8 + 9x) = 4(9x + 8)$

This is referred to as Commutative Property of Addition

From Line 2 to 3:

The property here depicts $a(b + c) = a * b + b * c$

i.e.

$4(9x + 8) = 4 * 9x + 4 * 8$

$4(9x + 8) = 36x + 32$

This is referred to as Distributive property

7 0

3 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

The ratio of the number of male lions to female lions in the animal reserve is 21:20. If there are 123 lions in the animal reser

Vanyuwa [196]

21 + 20 = 43

Total is 123, so divide by 43.

123/43 = 2.86 rounded to 3.

(females) 20 x 3 = 60 female lions

7 0

3 years ago