Eliot and his friends ate 3/8 of a pie. If the pie was cut into eight pieces ,how much pie is left over

At 8am,Ming Wei started travelling from Town A to Town B.At 8:40am, Ali started travelling from Town B to Town A.Mingwei's speed

Vesnalui [34]

Let's write some equations.

Mingwei's distance from Town A after $x$ hours from 8:00 is $45x$ .

Ali's distance from Town B after $x$ hours is $30x-20$ , since he doesn't start walking for 40 minutes.

When Mingwei's distance is twice Ali's, they've met up (since their distance from Town A is twice their distance from Town B).

So, this gives $45x=60x-40$ , so $15x=40$ , so $x=\frac{40}{15} = \frac{8}{3}$ , so the time is 10:40.

After $\frac83$ hours, Mingwei has traveled $45 * 8 / 3 = 120$ kilometers while Ali has traveled sixty, so the distance between the towns is $180$ kilometers.

7 0

2 years ago

.<br> 1.<br> f(x) = -2lx- 3| +1<br><br> How do I graph this?

ira [324]

Answer:

See explanation

Step-by-step explanation:

f(x) = -2lx-3| +1

1.lx-3|==>Translate the basic absolute value graph f(x)=|x| 3 units to the right

2. -lx-3| ==> Flip the graph over the x-axis

3. -2lx-3| ==> Double all the y-coordinates of the graph

4. -2lx-3| +1 ==> Translate the graph 1 unit up

4 0

1 year ago

Read 2 more answers

The chef has 80 pounds of strip loin. Th

PIT_PIT [208]

Um i was thinking 5% or something like that

5 0

2 years ago

Read 2 more answers

Question points)

mash [69]

Answer:

listen to lil peep ghost girl and I'll answer this for you

Step-by-step explanation:

just playing with u i tried but i can't find the darn answer

5 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