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3 years ago

5

Please help I will give brainliest only to those who answer respectfully and the best answer

1 answer:

katrin2010 [14]3 years ago

5 0

Answer:

possibly you could use kid per candy, number of trickery treaters or something

you could explain lets say over the holidays Miss. So and So's class's kids got an average of around 4-6 Christmas presents while Miss. Mary Sue's class had kids with 2 presents and kids with 12, since the range between 2-12 is bigger it had a greater standard of deviation

(hopefully this helps and I am not overcomplicating or incorrecting the standard of deviation, good luck)

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3 years ago

How many ways can a cook arrange 9 of her 12 spice jars on a spice rack?

pshichka [43]

It depends if the 12 spice jars are all different.

If they are:

_ _ _ _ _ _ _ _ _ (9 spaces)

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5 0

3 years ago

Read 2 more answers

Kay has 28 coins which includes nickels and dimes if the total is 2.35$, how many of each coin does Kay have?

m_a_m_a [10]

Answer:9 nickels and 19 dimes.

Step-by-step explanation:

Let $n$ be the number of nickels Kay has.

Let $d$ be the number of dimes Kay has.

Value of $1$ nickel is $ $0.05$

Value of $n$ nickels is $ $0.05n$

Value of $1$ dime is $ $0.1$

Value of $d$ nickels is $ $0.1dn$

Given that Kay has a total of $28$ coins.

So, $n+d=28$ ...(i)

Given that Kay has a total value of $ $2.35$

So, $0.05n+0.1d=2.35$ ...(ii)

Using (i) and (ii),

$0.05n+0.1(28-n)=2.35\\0.05n+2.8-0.1n=2.35\\2.8-2.35=0.05n\\0.45=0.05n\\n=9$

$d=28-n=28-9=19$

4 0

3 years ago

Calculate the density of a material that has a mass of 100 g and a volume of 25<br> cm 3 .

irakobra [83]

4g/cm^3

D = M/V
100g/25cm^3 = 4g/cm^3

5 0

3 years ago

g A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 13

Luden [163]

Answer:

There is enough evidence to reject the manufacturer’s claim and the standardized test statistic z is -3.43

Step-by-step explanation:

A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135°F.

So, Null hypothesis: $H_0:\mu \geq 135$

Alternate hypothesis : $H_a:\mu$

To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133°F.

x=133

n = 32

Population Standard deviation = $\sigma = 3.3^{\circ}F$

Formula : $z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

$z=\frac{133-135}{\frac{3.3}{\sqrt{32}}}\\\\z=-3.428$

z=-3.43

Refer the z table for p value

p value = 0.0003

$\alpha = 0.10$

p value < α

So, we failed to accept null hypothesis

Hence there is enough evidence to reject the manufacturer’s claim and the standardized test statistic z is -3.43

6 0

3 years ago