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scoray [572]
3 years ago
14

Please answer i will mark as brainliest

Mathematics
1 answer:
devlian [24]3 years ago
6 0
Use Photomath it would tell you the answer if you take a pic of your question
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Please solve these two questions :) - Use the graphs to solve the equations provided.
harkovskaia [24]

Answer:

7.(1,0) and (-4,0)

8.  (-3,0) and (-2,0)

Step-by-step explanation:

7.

When the function is equal to zero, we are looking for the zeros, or the x intercepts

Where it crosses the x axis are 1 and -4

(1,0) and (-4,0)

8.  When the function is equal to zero, we are looking for the zeros, or the x intercepts

Where it crosses the x axis are -3 and -2

(-3,0) and (-2,0)

8 0
3 years ago
Read 2 more answers
How do you solve Ax+3y=48 for A?
olga_2 [115]
A=-3y/x+48/x
Hope this helps
3 0
3 years ago
Write each percent as a fraction in simplest form. .45 1/2%​
Ann [662]
.45 is 45/100 (simplify) 9/20

1/2 is already simplest form of fraction.
7 0
4 years ago
From chapter Lines and angles
kvv77 [185]
Z is 40 x is 15 and 4 is 40
4 0
3 years ago
Suppose 12% of students are veterans. From a sample of 263 students, how unusual would it be to have less than 22 veterans?
kkurt [141]

Answer:

Yes, both np and n(1-p) are ≥ 10

Mean = 0.12 ; Standard deviation = 0.02004

Yes. There is a less than 5% chance of this happening by random variation. 0.034839

Step-by-step explanation:

Given that :

p = 12% = 0.12 ;

Sample size, n = 263

np = 263 * 0.12 = 31.56

n(1 - p) = 263(1 - 0.12) = 263 * 0.88 = 231.44

According to the central limit theorem, distribution of sample proportion approximately follow normal distribution with mean of p = 0.12 and standard deviation sqrt(p*(1 - p)/n) = sqrt (0.12 *0.88)/n = sqrt(0.0004015) = 0.02004

Z = (x - mean) / standard deviation

x = 22 / 263 = 0.08365

Z = (0.08365 - 0.12) / 0.02004

Z = −1.813872

Z = - 1.814

P(Z < −1.814) = 0.034839 (Z probability calculator)

Yes, it is unusual

0.034 < 0.05 (Hence, There is a less than 5% chance of this happening by random variation.

7 0
3 years ago
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