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3 years ago

Please help for 20 POINTS I would appreciate it if you put an answer you actually think is right :)

Mathematics

2 answers:

Oliga [24]3 years ago

8 0

It’s x!!!!!!!!!!!! !!!!!

Elina [12.6K]3 years ago

4 0

Answer: i would say that i think its x

Step-by-step explanation:

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Beto is selling raffle tickets to raise money for his school band. The tickets cost $12 each. Beto's goal is to make at least $2

Kamila [148]

Answer:

12x + 96 ≥ 200

Step-by-step explanation:

Each ticket costs 12 each= 12x

He has already sold 96=+96

he needs to make 200

12x+96>= 200

Plz mark brainliest

4 0

3 years ago

I forget how to do this help I'll give brainliest

Gnoma [55]

50 since a triangle has to equal 180, trust!

6 0

3 years ago

Convert to decimal 5 3/7

ASHA 777 [7]

5 3/7<span> = 5.428571428571429 or 5.43</span>

8 0

3 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago

Sterling silver consists of 92.5% silver and 7.5% copper. What decimal represent the portion of silver in Sterling silver?

mr Goodwill [35]

You have to move the decimal point two places to the left to convert a percent to a decimal, so the answer is 0.925.

6 0

3 years ago