Solve |2x- 6|<10 O XI-2 O [xl-8 O x12

Israel and Carlos were both eating at the same restaurant. Israel bought 4 hamburgers and 2 orders of fries for $9.74. Carlos bo

Likurg_2 [28]

Let
x-----------> cost of one hamburger
y-----------> cost of one order of fries

we know that
4x+2y=9.74--------------> equation 1
12x+y=22.77-----------> y=22.77-12x-----------> equation 2

using a graph tool---------> calculate the equations system
see the attached figure

the solution is
x=1.79
y=1.29

the cost of one hamburger is $1.79
the cost of one orders of fries is $1.29

If Tashiana buys 3 hamburgers and one order of fries
3*1.79+1.29---------> $6.66

the answer is $6.66

4 0

4 years ago

Choose a number sentence is that illustrates distribute property of multiplication over addition 2×7+8 = 2+7×2+8

Agata [3.3K]

Answer:

$(b)\ 2 * (7 + 8) = (2 * 7) + (2 * 8)$

Step-by-step explanation:

Given

$(a)\ 2 * (7 + 8) = (2 + 7) * (2 + 8)$

$(b)\ 2 * (7 + 8) = (2 * 7) + (2 * 8)$

$(c)\ 2 * (7 + 8) = (2 * 7) + 8$

Required

Which illustrates distributive property of multiplication over addition

The property states that:

$a(b + c) = a * b + a * c$

By comparison

$a = 2\ b = 7\ c = 8$

So, we have"

$2(7 + 8) = 2 * 7 + 2 * 8$

$2(7 + 8) = (2 * 7) + (2 * 8)$

4 0

3 years ago

Estimate the measure of the following angles

DIA [1.3K]

Answer:

For A I'd say about 100° and for B about 150° (This is just my estimation)

Step-by-step explanation:

A is almost a right angle but a little more than 90°.

B is almost a straight/flat angle but it's less than 180°.

7 0

3 years ago

Read 2 more answers

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

$Z = \frac{X - \mu}{s}$

X = 205

$Z = \frac{X - \mu}{s}$

$Z = \frac{205 - 200}{5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{195 - 200}{5}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

$Z = \frac{X - \mu}{s}$

$Z = \frac{210 - 200}{5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772.

X = 195

$Z = \frac{X - \mu}{s}$

$Z = \frac{190 - 200}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0

3 years ago

After 5 years, Rachel’s amount in her savings account is $7446.52. Her account earned interest at 4.32% APR and was compounded d

Katyanochek1 [597]

$\bf \qquad \textit{Continuously Compounding Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &7446.52\\
P=\textit{original amount deposited}\\
r=rate\to 4.32\%\to \frac{4.32}{100}\to &0.0432\\
t=years\to &5
\end{cases}
\\\\\\
7446.52=Pe^{0.0432\cdot 5}\implies \cfrac{7446.52}{e^{0.0432\cdot 5}}=P\implies 5999.92404\approx P$

5 0

3 years ago