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lyudmila [28]
3 years ago
6

$18.75 to $18.60 identify the percent of change as an increase or decrease . then find the percent of change round to the neares

t tenth of a percent , if necessary
Mathematics
2 answers:
Vilka [71]3 years ago
7 0

Answer:

0.8%

Step-by-step explanation:

mylen [45]3 years ago
6 0
0.8 I think it’s the answer
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Input Output Table 2 Input Output 12 19 13 20 14 ? 15 22 16 23 Which rule describes the relationship between the input/output va
Paha777 [63]
Each output has a value of 3 more than its input.

an = a(n-1) + 3

Missing number is 14 + 3  = 17.
5 0
3 years ago
Katherine is younger than Alexandra. Their ages are consecutive integers.
Serjik [45]

Using equations to solve the word problem, Katherine's age is 8 years

<h3>Word Problem</h3>

Word problem are often mathematical problem that are represented in words and sometimes can be solved using an equation.

To solve this problem, we have to represent the problem using mathematical equations;

Let;

Katherine's age = x²

Alexandra = 5(x + 1)

Sum of their age = 109

5(x + 1) + x² = 109

5x + 5 + x² = 109

x² + 5x - 104 = 0

Solving for the equation; x = 8

Katherine's age is 8

Learn more on word problem here;

brainly.com/question/13818690

#SPJ1

7 0
1 year ago
Professor Livingstone loves to perform acts of kindness. He decides to conduct an experiment in which he performs an act of kind
lana66690 [7]

Answer:

im sorry but im not going to read all that

Step-by-step explanation:

3 0
2 years ago
A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0
2 years ago
Can someone pls help me<br> Pls
lora16 [44]

Answer:

help u with what? Say what u need

5 0
2 years ago
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