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egoroff_w [7]
3 years ago
6

Please help I’ll give brainlyist

Mathematics
2 answers:
lakkis [162]3 years ago
6 0

Answer:

The second choice

Step-by-step explanation:

kati45 [8]3 years ago
3 0

Answer:

+ 6x^{2} and -10y^{2}

Step-by-step explanation:

3x^{2} + 5y^{2} {} + 3 {} + 4y^{2} + 6 = 9x^{2} - y^{2}+ 9

3x^{2} + 9y^{2} {} + 9 {}  = 9x^{2} -  y^{2}+ 9

+ 6x^{2}    -10y^{2}

3x^{2} + 9y^{2} + 6x^{2} + 9 - 10y^{2} = 9x^{2} - y^{2}+ 9

9x^{2} - y^{2}+ 9 = 9x^{2} - y^{2}+ 9

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Write a real world problem using ratios 5 to 9 and 12 to 15.
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Write the ratios in different formats EX: a to b, a:b, and a/b
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The box plots below show student grades on the most recent exam compared to overall grades in the class:
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Read 2 more answers
What is 14c12?<br> A. 78<br> B. 91<br> C. 105<br> D. 120
Delicious77 [7]

9514 1404 393

Answer:

  B.  91

Step-by-step explanation:

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3 years ago
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).

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3 years ago
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gladu [14]
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