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3 years ago

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The scale of a map says that 8 cm represents 5 km. What is the actual number of kilometers that is represented by 5 centimeters

on the map? Must show your math thinking on the sketchpad. You may use a calculator, but must show the calculations you used.

1 answer:

mel-nik [20]3 years ago

4 0

I don’t even know man but I’ll look into it for you

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Rosa and her friends are eating out for dinner. the bill was $30.12. they want to leave a 19% tip.

FromTheMoon [43]

Answer:

35.84

Step-by-step explanation:

You would take 19% of 30.12$ by doing 30.12x0.19 to get 5.72.

you then add 5.72 to 30.12 to get your total of 35.84$

7 0

2 years ago

Which shows one way to determine the factors of x3 11x2 – 3x – 33 by grouping? x2(x 11) 3(x – 11) x2(x – 11) – 3(x – 11) x2(x 11

Aloiza [94]

The one way to determine factors of x³ + 11x² – 3x – 33 will be $x^2(x+11)-3(x+11)_$ .

<h3>What is a factorization?</h3>

It is the method to separate the polynomial into parts and the parts will be in multiplication. And the value of the polynomial at this point will be zero.

The steps involved in the factorization are;

1. For each pair of parentheses, we create a common factor.

2. We use x2 as a common factor for the first parenthesis.

3. We use common factor 3 for the second parenthesis.

$x^3 + 11x^2 - 3x -33 \\\\ x^2(x+11)-3(x+11)_$

We will find the final solution as $x^2(x+11)-3(x+11)_$ .

Hence the one way to determine factors of x³ + 11x² – 3x – 33 will be $x^2(x+11)-3(x+11)_$ .

To learn more about the factorization refer to the link;

brainly.com/question/24182713

3 0

2 years ago

Given h(x) = 4x2 + 3x – 8, what is the value of h(–9) ?

fenix001 [56]

Answer:

h(-9) = -107

Step-by-step explanation:

h(-9) = 4(-9)2 + 3(-9) - 8 = -107

6 0

2 years ago

Evaluate each logarithm. Do not use a calculator. Log (125)(1/25)

svetlana [45]

Answer:

-2/3

Step-by-step explanation:

Log125(1/25)=x

Raise each side to the base of 125

125^Log125(1/25)=125^x

1/25 = 125^x

Rewrite 25 as a power of 5 and 125 as a power of 5

1 / 5^2 = 5^3^x

The if power is in the denominator, we can bring it to the numerator by making it negative

5^-2 = 5^3^x

We know that a^b^c = a^(b*c)

5^-2 = 5^(3*x)

Since the bases are the same, the exponents are the same

-2 = 3x

Divide by 3

-2/3 = 3x/3

-2/3 =x

6 0

3 years ago

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% 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 nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

<em>Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>Part I</u> : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

<u />

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

<u>Part II</u> : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% 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 nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

<em>Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

<u />

<u>98% confidence interval for</u> $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

2 years ago