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Sphinxa [80]
2 years ago
15

PLZ HELP ME WITH THIS!!! I FORGOT MY NOTESSS its due in 2 days

Mathematics
2 answers:
MAXImum [283]2 years ago
6 0
The picture is a black screen, so ami the problem?? Uduxnejcjdj
Aliun [14]2 years ago
5 0
You don't have a picture ?¿
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Ms. Martinez bought 36-pint sized containers of strawberry yogurt. How many gallons is that?
BabaBlast [244]

Answer:

4.5 gallons

Step-by-step explanation:

6 0
2 years ago
Solve the system of equations algebraically. <br> 6x + 5y = -14<br> 5x + 2y = -3
Alik [6]
(1,-4)
x=1, y=-4.

Solve for the first variable in one of the equations, then substitute the result into the other equation.
5 0
2 years ago
genetic experiment with peas resulted in one sample of offspring that consisted of green peas and yellow peas. a. Construct a ​%
Andreas93 [3]

Complete Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 432 green peas and 164 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?

Answer:

The  95%  confidence interval is  0.2392  <  p < 0.3108

No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%

Step-by-step explanation:

From the question we are told that

  The total sample size is  n  =  432 + 164 =596

   The  number of  offspring that is yellow peas is y =  432

   The  number of  offspring that is green peas   is g =  164

   

The sample proportion for offspring that are yellow peas is mathematically evaluated as

        \r p  =  \frac{ 164 }{596}

        \r p  =  0.275

Given the the  confidence level is  95% then the level of significance is mathematically represented as

       \alpha  =  (100 - 95)\%

      \alpha =  5\%  =  0.0 5

The  critical value of  \frac{\alpha }{2} from the normal distribution table is  

      Z_{\frac{\alpha }{2} } = 1.96

Generally the margin of error is mathematically evaluated as

        E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1- \r p )}{n} }

=>      E = 1.96 * \sqrt{\frac{0.275 (1- 0.275 )}{596} }

=>      E =  0.0358

The  95%  confidence interval is mathematically represented as

      \r p - E  <  p < \r p + E

=>   0.275 -  0.0358  <  p < 0.275 +  0.0358

=>   0.2392  <  p < 0.3108

6 0
3 years ago
The opening balance of the March billing cycle for Bernice's credit card was $2374. If she makes a new purchase of $200 on the 2
Norma-Jean [14]

The opening balance of the March billing cycle for Bernice's credit card = $2374.

Amount of new purchase = $200 on the 20th of March.

She did purchase only of $200 in that March month.

We are given that she didn't do any payment.

So, still she has balance $2374 till end of the month.

Therefore, her average daily balance is $2374.00

So, the correct option is  B. $2374.00.

4 0
3 years ago
Read 2 more answers
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quester [9]
First substitute, then distribute, then add, then simply
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