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3 years ago

Classify the expression below.

Mathematics

1 answer:

babymother [125]3 years ago

5 0

Answer:

Step-by-step explanation:

It's a quartic trinomial because its has a fourth degree variable in the expression and only 3 variables in the equation.

2x⁴ - x + 5.

Linear trinomials have 1 degree 3 variables in the monomial. Eg x + y + 2 = 0 is an example of a linear trinomial.

Cubic trinomials have 3 degree and 3 variables in the equation eg = 5x³ + 2x + 1 = 0

Quadratic trinomials is the most common trinomial we deal with everyday eg Ax² + bx + c = 0

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What is the equation of the line

LenKa [72]

Answer:

y=0

Step-by-step explanation:

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2 years ago

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A sauce recipe uses 4 cans of tomatoes in every 2/12 quarts. How many cans of tomatoes are used in each quart of the sauce?

My name is Ann [436]

4/12 quarts
hope this helps

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3 years ago

How do I solve this problem

guapka [62]

You first subtract 1/5 from both sides
3/4-1/5 = 11/20
3/2x=11/20

now you multiply 2/3 on both sides since you have to get x

11/20 * 2/3
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3 years ago

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:

$\alpha =1-0.98=0.02$

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3 years ago

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In Sunday she babysits 1 hour and 23 minutes,in Saturday she babysit 1 hour and 39 minutes.In total it’s 3 hours and 12 minutes.

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2 years ago