Use the table to identify the values of p and q that should be used to factor x2

You are making snap bags, you want each bag to have the same combination you have 75 carrots and 40 celery sticks and you want t

makkiz [27]

Answer:

5 Bags

Step-by-step explanation:

There are 75 carrots and 40 celery sticks.

If we want each bag to have the same combination, the greatest number of bags we can make is determined by the Greatest common divisor of the two numbers.

To find the Greatest common divisor of 70 and 40

Step 1: Express each number as a product of its prime factors

40=2 X 2 X 2 X 5
75=3 X 5 X 5

Step 2: Pick each pair of common factors.

40=2 X 2 X 2 X 5
75=3 X 5 X 5

The only factor common to both pairs is 5.

Therefore, the Greatest common divisor of 70 and 40 is 5.

The greatest number of bags that can be made is 5.

4 0

3 years ago

The radius of a circle is 4 1/2 inches. What is the circles circumference in terms of 3.14 (pie)

DerKrebs [107]

The circumference of a circle can be found using the equation 2 π r

(2)(3.14)(4.5) = 28.26 (or 28 13/50 if you need it in fractions)

3 0

3 years ago

Suppose you went to a carnival. The price to get in was $6, and you paid $0.50 to get on each ride. If you went to the carnival

kondor19780726 [428]

11.50$
.50 x 11
=5.50 + 6
=11.50

8 0

3 years ago

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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

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

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

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

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

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

3 years ago

A line of best fit was drawn for 5 data points. what is the maximum number of these data points that may not actually be on the

jenyasd209 [6]

[b.3] ^_^
hopefully i helped (0-0)

4 0

3 years ago

Read 2 more answers