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4 years ago

6

Caleb bought groceries and paid \$1.60$1.60dollar sign, 1, point, 60 in sales tax. The sales tax rate is 2.5\%2.5%2, point, 5, p

ercent. What was the price of Caleb's groceries, before tax?

1 answer:

saveliy_v [14]4 years ago

5 0

Answer:

Price of Caleb's groceries before tax = $64

Step-by-step explanation:

Let the price of groceries before tax be =$ $x$

Sales tax charged = $1.60

Sales tax rate =2.5%

Sales tax charged in terms of $x$ will be = 2.5% of the Original price of grocery = $2.5\% of x = 0.025\times x=0.025x$

So, we have,

$0.025x=1.60$

Dividing both sides by $0.025$

$\frac{0.025x}{0.025}=\frac{1.60}{0.025}$

∴ $x=64$

∴ Price of Caleb's groceries before tax = $64

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Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has

Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= <u> X[bar] - μ </u>~ N(0;1)

σ/√n

The formula for the interval is:

X[bar] ± $Z_{1- \alpha /2}$ * (σ/√n)

$Z_{1- \alpha /2}= Z_{0.90} = 1.28$

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= $\frac{(3.26-3.04)}{2}$ = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= $Z_{1- \alpha /2}$ * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= $Z_{1- \alpha /2}$ * (σ/√n)

$\frac{d}{Z_{1- \alpha /2}}$ = σ/√n

√n * $\frac{d}{Z_{1- \alpha /2}}$ = σ

√n = σ * $\frac{Z_1- \alpha /2}{d}$

n = (σ * $\frac{Z_1- \alpha /2}{d}$ )²

n= (0.33 * $\frac{1.28}{0.08}$ )²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

4 0

3 years ago

PLEASE HELP MEH!

goblinko [34]

Answer: x=72, y=26

Step-by-step explanation:

Since a line is 180°, we can do:

2x+10+y=180

2x+y=170

Since a triangle is 180° also, we can do:

x+y+82=180

x+y=98

If we subtract x+y=98 from 2x+y=170, we get:

2x+y=170

x+y=98

x=72

If we insert x=72 into x+y=98, we get:

72+y=98

y=26

We can check our answer by inserting x=72 and y=26 into x+y+82=180

72+26+82=180

98+82=180

180=180

7 0

4 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$

8 0

3 years ago

Which scenario best matches the inequality x>3?

jek_recluse [69]

The correct answer is B.

8 0

3 years ago

Read 2 more answers

Parking lot has 121 spots. Ratio of senior to junior is 7 to 4. How many spots go to each grade level?

Andrej [43]

I am going to build a chart this is best when using ratios

  seniors junior total
7 4
*? *? *?
Totals 121

The way we fill out this chart is that we add the rows and multiply the columns.

So we add 7 + 4 and fill it in our chart we get 11

  seniors junior total
7 4 11
*? *? *?
Totals 121

now to find the *? need to divide 121/11 to get *?

121/11 = 11

so now our chart looks like this
seniors junior total
7 4 11
11 11 11
Totals 121

Now we multiply each column

so
7 * 11 = 77
4*11 = 44

now our chart look like this

seniors junior total
7 4 11
11 11 11
Totals 77 44 121

so seniors get 77 spaces
and juniors get 44 spaces.

6 0

3 years ago