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Svetach [21]
3 years ago
8

Will give points and brainliest

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0
Where is the picture I so i can anwser
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A notebook cost $1.50 and a binder cost $6.50 jesica bought x notebooks and y binders
Bond [772]
1.50x + 6.50y is your algebraic expression.
4 0
3 years ago
Read 2 more answers
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 436.0 gram setting. It
Vilka [71]

Answer:

We conclude that the machine is under filling the bags.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 436.0 gram

Sample mean, \bar{x} = 429.0 grams

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

H_{0}: \mu = 436.0\text{ grams}\\H_A: \mu < 436.0\text{ grams}

We use one-tailed(left) z test to perform this hypothesis.

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{429 - 436}{\frac{23}{\sqrt{40}} } = -1.92

Now, z_{critical} \text{ at 0.05 level of significance } = -1.64

Since,  

z_{stat} < z_{critical}

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that the machine is under filling the bags.

8 0
3 years ago
What is the answer 6v(2v + 3)
Firlakuza [10]
See attached picture


Hope this helps

6 0
3 years ago
Read 2 more answers
Atul has 2/3 lb of candy. Jose has 3/5 lb and Maria has 1/2 lb less then José. How many more pounds of candy does Atul have than
liraira [26]

Given:

Atul has \dfrac{2}{3} lb of candy.

Jose has \dfrac{3}{5} lb of candy.

Maria has \dfrac{1}{2} lb less than Jose.

To find:

How many more pounds of candy does Atul have than Maria?

Solution:

Since, Maria has \dfrac{1}{2} lb less than Jose, therefore

Maria has = \dfrac{3}{5}-\dfrac{1}{2} lb

Maria has = \dfrac{6-5}{10} lb

Maria has = \dfrac{1}{10} lb

Difference between the candies Atul and Maria have.

Difference = \dfrac{2}{3}-\dfrac{1}{10} lb

Difference = \dfrac{20-3}{30} lb

Difference = \dfrac{17}{30} lb

Therefore,  Atul have \dfrac{17}{30} lb of candy more than Maria.

6 0
3 years ago
Please help me with this sum​
Scilla [17]

Answer:

1. 384

2. 180

3. 324

4. 648

5. 476

6. 528

7. 294

8. 384

9. 414

10. 80

11. 72

12. 301

13. 222

14. 146

15. 156

16. 270

17. 283

18. 300

19. 292

20. 58

21. 64

22. 138

23. 296

24. 170

Hope it helps you

7 0
2 years ago
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