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3 years ago

5

Please give me brainly! i really wanna level up and I need 5 ! i will not force but please do!

1 answer:

jasenka [17]3 years ago

4 0

Answer:

ill ask a question if u want

Step-by-step explanation:

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Mary and Jeff both have jobs at a baseball stadium selling bags of peanuts. They each get paid $12 per game

Anarel [89]

A) 1.75p + 12 = 54
1.75p = 42
p = 24 bags

b) 1.75(70) + 12 = b
122.5 + 12 = b
b = $134.50

6 0

2 years ago

PLS HELP!!! WILL GIVE BRAINLIEST!!

Tasya [4]

Answer:

?

Step-by-step explanation:

?

3 0

2 years ago

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If the public debt of the United States in 2004 was $7,379,052,696,330.32 and the budget deficit in 2005 was $553,656,965,393.18

slava [35]

Since there is a shortage in the budget in 2005 by 553,656,965,393.18, the total debt that of the United States in 2005 will be the public debt by 2004 added with this deficit in the budget. Having said that, the public debt by 2004 should be equal to,
$ 7.9327 x 10^12

3 0

3 years ago

Read 2 more answers

I attached the question

noname [10]

5x = 4y - 3

The correct answer is D. 5x - 4y = -3

This is because standard form is the form ax + by = c, where a, b, and c are coefficients or constants of the variables x and y. Thus, Answer D (above) is the correct answer to this problem, as it is the only option that fits this.

4 0

3 years ago

The average weight of the entire batch of the boxes of cereal filled today was 20.5 ounces. A random sample of four boxes was se

zhenek [66]

Answer:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s= 0.286$

And then the estimator for the standard error is given by:

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

Step-by-step explanation:

For this case we have the following dataset given:

20.05, 20.56, 20.72, and 20.43

We can assume that the distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for this case would be:

$SE= \frac{\sigma}{\sqrt{n}}$

And we can estimate the deviation with the sample deviation:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s= 0.286$

And then the estimator for the standard error is given by:

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

4 0

3 years ago