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2 years ago

A store sells 12 cans of soup for $21. How much would it cost you to buy 5 cans of soup?

Mathematics

1 answer:

saw5 [17]2 years ago

7 0

Answer:

8.75$

Step-by-step explanation:

if 12 cans cost $21

1 can cost $21÷12=1.75$

5 cans = 1.75$ ×5

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A pair of shoes cost $120 but they are 25% Off;

Sever21 [200]

Answer:

Discount = $30, Final price = $90

Step-by-step explanation:

25% is one quarter. 120/4=30

One quarter of 120 is 30, so - 30 would give discounted price of $90.

Similarly you could think of it like the discounted price is 75% of the original one and do 120x0.75=90 to find 75% of 120.

Hope this helped!

3 0

3 years ago

7,000lbs times what would equal 3.5 tons

PSYCHO15rus [73]

7,000lbs x 1=3.5 tons. 3.5 tons is 7,000lbs.

6 0

3 years ago

1, -5, 25, -125,...

nadezda [96]

The answer I believe is the third choice.
3, -2, -7, -12
These all fit into f(n)-5=f(n+1)

8 0

3 years ago

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]