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Leto [7]
3 years ago
13

Janelle would like to give a party bag to every person who is coming to her party. The cost of the party bag is $7 per person. W

rite an inequality that describes the number of people P that she can invite if Janelle has D dollars to spend on the party bags.
Mathematics
1 answer:
torisob [31]3 years ago
7 0
D >7d. The number of dollars must be
- greater than or equal to 7 times the
number of people because she can't spend more than she has.
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F(x) = 8|x| - 5<br> f(-2) =
Anastaziya [24]

Answer:

f(-2) = 11

Step-by-step explanation:

Hello!

We can evaluate y by plugging in -2 for x in the equation.

<h3>Evaluate</h3>
  • f(x) = 8|x| - 5
  • f(-2) = 8|-2| - 5
  • f(-2) = 8(2) - 5         Remember that absolute value turns the number positive
  • f(-2) = 16 - 5
  • f(-2) = 11

The evaluated value is 11.

6 0
2 years ago
PLS EXPLAIN IN FULL DETAIL
elena-14-01-66 [18.8K]

Answer:

1.50x=18

x=12

Step-by-step explanation:

You have to divide 1.50 on both sides and 1.50x/1.50= x and 18/1.50=12 so x=12

3 0
3 years ago
Read 2 more answers
A person invest $10,000 into a bank the bank pays 4.75% interest compounded semi annually. To the nearest 10th of a year, how lo
Mama L [17]

Answer:

T is 13.9 years to the nearest 10th of a year

Step-by-step explanation:

In this question, we are to calculate the number of years at which someone who invests a particular amount will have a particular amount based on compound interest.

To calculate the number of years, what we do is to use the compound interest formula.

Mathematically,

A = P(1+ r/n) ^nt

Where A is the final amount after compounding all interests which is $19,200 according to the question

P is the initial amount invested which is $10,000 according to the question

r is the rate which is 4.75% according to the question = 4.75/100 = 0.0475

n is the number of times per year in which interest is compounded. This is 2 as interest is compounded semi-annually

t= ?

we plug these values;

19200 = 10,000(1+0.0475/2)^2t

divide through by 10,000

1.92 = (1+0.02375)^2t

1.92 = (1.02375)^2t

We find the log of both sides

log 1.92 = log [(1.02375)^2t)

log 1.92 = 2tlog 1.02375

2t = log 1.92/log 1.02375

2t = 27.79

t = 27.79/2

t = 13.89 years

The question asks to give answer to the nearest tenth of a year and thus t = 13.9 years

7 0
3 years ago
Thanks, brainliest, and 5 stars!!
frez [133]
I hope this helps its step 2 gl
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
4 years ago
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