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svp [43]
3 years ago
8

PLEASE HURRY i really need help with this

Mathematics
1 answer:
Serjik [45]3 years ago
4 0
180-45= 135, the answer is 135
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On average, Haley is able to read 4 5/6 Chapters in the her book each week . How many weeks will it take Haley to read 9 2/3 Cha
Step2247 [10]

Answer:

2 weeks

Step-by-step explanation:

y = mx + b

y = total number of chapters read = 9\frac{2}{3}

m = number of chapters read per week = 4\frac{5}{6}

x = number of weeks = what we're solving for

b is not needed to find this solution

9\frac{2}{3} = 4\frac{5}{6} x                            divide 4\frac{5}{6} from both sides

x = 2 weeks

6 0
3 years ago
Solve for x<br> 2800 = 400.210x
dusya [7]

Answer:

2800=400.21x

x = 6.996327

3 0
3 years ago
Please help with this question, giving brainliest
BartSMP [9]
Answer: B
Work:
-5x+10>-15 = x<3/10
3 0
2 years ago
Please help me ASAP. For homework due right now
sveta [45]

Answer:

at least 20 rows.

Step-by-step explanation:

when you add the students, teachers, and chaperones, it comes to 159. then you divide 159 by the number of seats each row has, so 159/8 = 19.875, you would have to round to the nearest whole number so 20 should be correct.

5 0
3 years ago
US average math SAT scores follow a normal distribution with a mean of 505 and a standard deviation of 112. A sample of 64 enter
Hitman42 [59]

Answer:

The claim that the scores of UT students are less than the US average is wrong

Step-by-step explanation:

Given : Sample size = 64

           Standard deviation = 112

           Mean = 505

           Average score = 477

To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.

Solution:

Sample size = 64

n > 30

So we will use z test

H_0:\mu \geq 477\\H_a:\mu < 477

Formula : z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

z=\frac{505-477}{\frac{112}{\sqrt{64}}}

z=2

Refer the z table for p value

p value = 0.9772

α=0.05

p value > α

So, we accept the null hypothesis

Hence The claim that the scores of UT students are less than the US average is wrong

3 0
3 years ago
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