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levacccp [35]
3 years ago
15

Factor the expression using the GCF. 100-80 = it’s not 20 btw Plz help ASAP

Mathematics
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

20

Step-by-step explanation:

(are you sure it isnt? i think it is 20 )

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Simplify 5 2 thirds
erica [24]
Answer= 2/3
5 2/3= multiply 5x3=15, then 5x2=10 which makes the fraction: 10/15, simplify by finding the highest common denominator that divides 10 & 15 equally, which would be 5,
5 divided into 10 is 2. 5 devided into 15 is 3 which makes your answer, 2/3.
not 100%, but it makes sense to me...
8 0
3 years ago
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A merchant bought a table for ₹8250 and spent ₹175 on its transportation. He
liubo4ka [24]

Answer: Loss of 925

==============================================

Work Shown:

Total cost = 8250+175 = 8425

Total revenue = 7500

Profit = revenue - cost = 7500 - 8425 = -925

Profit = -925

The negative profit means the merchant lost money.

6 0
2 years ago
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24. If 2A = 3B = 4C then A:B:Can<br>is<br>O a 23:4<br>O b. 4:32<br>O c. 6:4:3<br>O d. 3:4:2​
katovenus [111]

Answer:

the answer is c. 6:4:3

Step-by-step explanation:

a=6; b=4; c=3

3 0
3 years ago
The 20 to 34 age groups are approximately normally distributed with µ = 110
Savatey [412]

Answer:

21.94% of people aged 20 to 34 have IQs between 125 and 150.

Step-by-step explanation:

<u>The complete question is:</u> Scores on the Wechsler Adult Intelligence Scale (a standard IQ test) for the 20 to 34 age group are approximately Normally distributed with  μ = 110  and  σ = 25.

What percent of people aged 20 to 34 have IQs between 125 and 150?

Let X = <u><em>Scores on the standard IQ test for the 20 to 34 age group</em></u>

SO, X ~ Normal(\mu=110,\sigma^{2} =25^{2})

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

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

where, \mu = population mean = 110

            \sigma = standard deviation = 25

Now, the percent of people aged 20 to 34 have IQs between 125 and 150 is given by = P(125 < X < 150) = P(X < 150) - P(X \leq 125)

       P(X < 150) = P( \frac{X-\mu}{\sigma} < \frac{150-110}{25} ) = P(Z < 1.60) = 0.9452

       P(X \leq 125) = P( \frac{X-\mu}{\sigma} \leq \frac{125-110}{25} ) = P(Z \leq 0.60) = 0.7258

The above probability is calculated by looking at the value of x = 1.60 and x = 0.60 which has an area of 0.9452 and 0.7258.

Therefore, P(125 < X < 150) = 0.9452 - 0.7258 = 0.2194 or 21.94%

7 0
4 years ago
A school chorus has 90 sixth-grade students and 75 seventh-grade students. The music director wants to make groups of performers
Tresset [83]

Answer:

The music director can form a total of 15 groups

6 Sixth grade students per group

5 Seventh grade students per group

Step-by-step explanation:

Sixth grade students = 90

Seventh grade students = 75

Find the highest common factor of 75 and 90

90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

75 = 1, 3, 5, 15, 25, 75

The highest common factor of 75 and 90 is 15

Therefore, the music director can form a total of 15 groups

Sixth grade students per group = 90 / 15

= 6 Sixth grade students per group

Seventh grade students per group = 75 / 15

= 5 Seventh grade students per group

3 0
3 years ago
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