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tatiyna
3 years ago
9

Kira, goran, & joe sent a total of 90 text messages during the weekend. Goran sent 5 fewer messages than Kira. Joe sent 3 ti

mes as many messages as kira. How many messages did they each send
Mathematics
1 answer:
MissTica3 years ago
4 0

Answer:

Kira sent 19 text messages, Goran sent 14  text messages  and Joe sent 57  text messages.

Step-by-step explanation:

Let be "k" the number of text messages that Kira sent during the weekend, "g" the number of text messages that Gora sent during the weekend and "j" the number of text messages that Joe sent during the weekend.

We know know that they sent a total of 90 text messages then:

k+g+j=90

Goran sent 5 fewer messages than Kira. This is:

g=k-5

And Joe sent 3 times as many messages as Kira. Then:

j=3k

The steps to solve this are:

- Substitute the second equation and the third equatio into the first equation  and then solve for "k":

k+(k-5)+(3k)=90\\\\5k-5=90\\\\5k=90+5\\\\k=\frac{95}{5}\\\\k=19

- Substitute this value into the second equation to find "g":

g=19-5=14

- Substitute the value of "k" into the third equation to find "j":

j=3(19)=57

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A: What is the smallest increment on the voltmeter
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3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

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3 years ago
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