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4 years ago

Ava ate 5 crackers, or 5% of the entire box. How many crackers were in the box

Mathematics

2 answers:

Elena-2011 [213]4 years ago

6 0

100 because if she ate 5 crackers and 5% that means every cracker is equal to one percent!

Nezavi [6.7K]4 years ago

4 0

Answer:

5x2 = 10

10%= 10 crackers

10x 10= 100 crackers

100 crackers in the box

Step-by-step explanation:

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A positive real number is 1 more than another. When -2 times the smaller is added to the square of the larger, the result is 33.

Naya [18.7K]

Answer:

The smaller number is 4√2 and the larger number is (4√2 + 1).

Step-by-step explanation:

Let the two numbers be x and y, where y is the larger of the two numbers.

Since y is the larger number, it is one more than the smaller number. So:

$y=x+1$

When negative two times the smaller is added to the square of the larger, the result is 33. In other words:

$-2x+y^2=33$

Substitute:

$-2x+(x+1)^2=33$

Solve for x. Square:

$-2x+(x^2+2x+1)=33$

Simplify:

$x^2+1=33$

Subtract one from both sides:

$x^2=32$

And take the square root of both sides:

$x=\pm\sqrt{32}=\pm 4\sqrt{2}$

Since y is positive, we can ignore the negative case. So, the smaller number is:

$x=4\sqrt{2}\approx5.66$

And the larger number is:

$y = 4\sqrt{2} + 1 \approx6.66$

3 0

3 years ago

What's the percent increase from 3.49 to 3.89

earnstyle [38]

Hi there,

3.89% - 3.49% = 0.40%

Hence, it increased by 0.40%

Hope this helps :)

8 0

3 years ago

Read 2 more answers

A sample of 1500 computer chips revealed that 37% of the chips fail in the first 1000 hours of their use. The company's promotio

andreyandreev [35.5K]

Answer:

$z=\frac{0.37 -0.39}{\sqrt{\frac{0.39(1-0.39)}{1500}}}=-1.59$

Step-by-step explanation:

Data given and notation

n=1500 represent the random sample taken

$\hat p=0.37$ estimated proportion of of chips fail in the first 100 hours

$p_o=0.39$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the actual percentage that fail is different from the stated percentage. Find the value of the test statistic.:

Null hypothesis: $p=0.39$

Alternative hypothesis: $p \neq 0.39$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .