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MAXImum [283]
3 years ago
9

A mixture of 30 pounds of candy sells for $1.10 a pound. The mixture consists of chocolates worth $1.50 a pound and chocolates w

orth 90¢ a pound. How many pounds of the $1.50 chocolate were used to make the mixture?
A)15
B)10
C)12
Mathematics
1 answer:
astra-53 [7]3 years ago
3 0
x-chocolate\ worth\ 1,50\$ \ per\ pound\\
y-chocolate\ worth\ 0,9\$\ per\ pound\\
\\\\
 \left \{ {{x+y=30} \atop {30*1,10=x*1,50+y*0,9}} \right.\\\\ \left \{ {{x=30-y} \atop {33=1,5x+0,9y}} \right. \\\\ Using \ substitution\ method:\\\\33=1,5(30-y)+0,9y\\\\
33=45-1,5y+0,9y\\\\ -12=-0,6y\ |:-0,6\\\\
y=20\\
x=10
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Find the volume of the cylinder. round to the nearest tenth.
vaieri [72.5K]

Answer:

3217.00

Step-by-step explanation:

V=πr^2h=π·8^2·16≈3216.99088

5 0
2 years ago
Read 2 more answers
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
2 years ago
Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl
amm1812

Answer:

1. P(s_1|I)=\frac{1}{11}

2. P(s_2|I)=\frac{8}{11}

3. P(s_3|I)=\frac{2}{11}

Step-by-step explanation:

Given information:

P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3

P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1

(1)

We need to find the value of P(s₁|I).

P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}

P(s_1|I)=\frac{0.015}{0.165}

P(s_1|I)=\frac{1}{11}

Therefore the value of P(s₁|I) is \frac{1}{11}.

(2)

We need to find the value of P(s₂|I).

P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}

P(s_2|I)=\frac{0.12}{0.165}

P(s_2|I)=\frac{8}{11}

Therefore the value of P(s₂|I) is \frac{8}{11}.

(3)

We need to find the value of P(s₃|I).

P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}

P(s_3|I)=\frac{0.03}{0.165}

P(s_3|I)=\frac{2}{11}

Therefore the value of P(s₃|I) is \frac{2}{11}.

4 0
2 years ago
Does (2, -1) make this a true statement? Why or why not?<br> 3x + 2y = 4
kvv77 [185]

Answer: Yes

Step-by-step explanation:

So a coordinate pair is always set up (x,y) so you plug the x term in the coordinate pair into the x in the equation and the y term in for the y. 3×2 + 2×-1. Multiply them together and you end up with 4.

3 0
3 years ago
2 (3x+1)=17-3 (x-17)
elena-14-01-66 [18.8K]
<h2>Answer:</h2>

x =  \frac{22}{3}

<h3>alternative form:</h3>

x = 7 \frac{1}{3}  \: or \: x = 7.33333333333

<h2>Step-by-step explanation:</h2>

2(3x + 1) = 17 - 3( x - 17) \\ distribute \\  \\ 6x + 2 = 17 - 3x  + 51 \\ add \: 3x \: from \: both \: sides \\  \\ 9x + 2 = 17  +  51 \\ add \: 17 \: to \: 51 \\  \\ 9x + 2 = 68 \\ subtract \: 2 \: from \: both \: sides \\  \\ 9x = 66 \\  divide \: both \: sides \: by \: 9 \\  \\ x =  \frac{22}{3}

3 0
3 years ago
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