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When a baby panda is born, it is 6cm in length. Every week after this, its size increases by 5 cm. Write an equation in slope in

Neporo4naja [7]

Answer:

56cm, y=5x+6

Step-by-step explanation:

The baby panda's initial or starting length is 5cm. This is the y-intercept or b value. We also know that the panda increases every week by 5 cm. This is the rate of change or slope. This is m. We can write in slope-intercept form y=mx+b.

y=5x+6

We substitute x=10. y=5(10)+6=56cm

5 0

3 years ago

What two numbers multiply to get -56 and add to get -1?

sukhopar [10]

It is -8 and 7.
-8 x 7 = -56.
-8 + 7 = -1

Hope This Helps You!
Good Luck Studying :)

4 0

3 years ago

Divide 1-k^3 by k-1

UNO [17]

Answer:

k 2 + k + 1

Step By Step:

Cancel the common factor.

Then.

k 2 + k +1 by 1

Gets you k2+k+1

7 0

3 years ago

Please help me !!

kherson [118]

The second answer is D) it will increase 8 times

7 0

3 years ago

Read 2 more answers

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