I am operations that undo each other like multiplication and divison

Erica runs 5.3 kilometers every morning. One day, she runs 0.95 kilometers and then stops to take a break. How much further does

Leviafan [203]

Answer:

1foot(ft.)1yard(yd.)1mile(mi.)===12inches(in.)3feet(ft.)5,280feet(ft.)

Step-by-step explanation:

8 0

3 years ago

Who can help me with this entire page?

trasher [3.6K]

Answer:

I'm not entirely sure what the question is but every thing under 12.5 is closer to 12, everything 12.5 and upwards like 12.87 would be closer to 13

8 0

2 years ago

Which measurement is closest to the volume of the cone in cubic inches? 6in 16 in

Nookie1986 [14]

Answer:

it problely 16 is close

Step-by-step explanation:

it dipens where the numeber is place at

4 0

3 years ago

The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de

blsea [12.9K]

Answer:

1. 15.87%

2. 6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

$P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

$\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8$

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

$P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 6) =0.0228 = 2.28\%$

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

$P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%$

84% of newborn babies weigh between 6.7 and 9.5 pounds.

7 0

3 years ago

The table below shows the functions to represent the sale f(t), in thousands, of four companies in t years: Company A B C D Sale

bazaltina [42]

1) Company A and C

2)Your answer is f(t) = 180(0.5)^t This is because the number is cut in half for every hour.

3)C 0 ≤ x ≤ 50 is the right answer because the starting time 9:05 is considered as zero and the 9:55 is the ending point which is considered as 50.Or simply the difference of both the times is the domain of the function.

5 0

3 years ago