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3 years ago

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The California Board of Education wants to know how 5th grade students are doing in math this year. They get a list of all 5th g

raders
in the state of California and look at the overall math grades of every 10th student on the list. Then they add the scores and divide to
get an average. What kind of sampling is this?

1 answer:

kolbaska11 [484]3 years ago

3 0

Answer: I think random sample

Step-by-step explanation:

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-3x^2-5x+9=0 what is the value for a,b,and c

Tju [1.3M]

Answer:

a = -3 b = -5 c = 9

Step-by-step explanation:

A quadratic function is written in the form

ax^2 +bx +c =0

-3x^2-5x+9=0

a = -3 b = -5 c = 9

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mr Goodwill [35]

Three pieces are left.

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Translate 4 units right and 3 units up.

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Answer: Hope this helps! Sorry if it looks a little messy. Have an amazing day!

Step-by-step explanation:

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3 years ago

Find the indicated nth partial sum of the arithmetic sequence.<br> −7, −3, 1, 5, . . ., n = 30

TiliK225 [7]

Step-by-step explanation:

Given the Arithmetic sequence

$-7, -3, 1, 5, . . .$

An arithmetic sequence has a constant difference d and is defined by

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Compute\:the\:differences\:of\:all\:the\:adjacent\:terms}:\quad \:d=a_{n+1}-a_n$

$-3-\left(-7\right)=4,\:\quad \:1-\left(-3\right)=4$

$\mathrm{The\:difference\:between\:all\:of\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}$

$d=4$

$\mathrm{The\:first\:element\:of\:the\:sequence\:is}$

$a_1=-7$

as

$a_n=a_1+\left(n-1\right)d$

$\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:$

$a_n=4\left(n-1\right)-7$ ∵ $d=4$

$\mathrm{Arithmetic\:sequence\:sum\:formula:}$

$n\left(a_1+\frac{d\left(n-1\right)}{2}\right)$

$\mathrm{Plug\:in\:the\:values:}$

$n=30,\:\space a_1=-7,\:\spaced=4$

$=30\left(-7+\frac{4\left(30-1\right)}{2}\right)$

$=30\left(58-7\right)$ ∵ $\frac{4\left(30-1\right)}{2}=58$

$=30\cdot \:51$

$=1530$ ∵ $\mathrm{Multiply\:the\:numbers:}\:30\cdot \:51=1530$

Therefore, the indicated nth partial sum of the arithmetic sequence is 1530.

ANOTHER METHOD

as

$a_n=4\left(n-1\right)-7$

n = 30

$\sum _{n=1}^{30}\:4\left(n-1\right)-7$

$=\sum _{n=1}^{30}4n-11$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \sum a_n+b_n=\sum a_n+\sum b_n$

$=\sum _{n=1}^{30}4n-\sum _{n=1}^{30}11$

as

$\sum _{n=1}^{30}4n=1860$

and

$\sum _{n=1}^{30}11=330$

so

$=1860-330$

$=1530$

7 0

4 years ago

In need of help PLEASE!!!!

ozzi

Answer:

a bc I think it's right idk really

Step-by-step explanation:

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7 0

3 years ago