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3 years ago

What is the difference in simplest terms of 3/4 - 1/3

Mathematics

1 answer:

GuDViN [60]3 years ago

6 0

Answer:

5/12

Step-by-step explanation:

can't be anymore simplified, it already is

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-UIVIT UUNNU INTEREST

Jobisdone [24]

Answer:

Step-by-step explanation:

Principal, P = $4,000

Interest rate, r = 3% = 0.03

Period, t = 20 years

Number of times compounded in a year = 4

Amount , A = P( 1 + r/n)^tn

A = 4000( 1 + 0.03/4)^4*20

A = 4000( 1 + 0.0075) ^80

A =4000( 1.0075) ^80

A = 4000* 1.818

A = $7272.18

5 0

3 years ago

Helppppppppppppppppppppppppppppppppppp

skad [1K]

Answer:

The answer would be 7 I hope this helps! : )

4 0

2 years ago

Read 2 more answers

The table shows the amounts of food collected by two homerooms. Homeroom A collects 21 additional items of food. How many more i

Andreyy89

<span>Let x be additional food item the homeroom B should collect to have greater item per student than homeroom A.</span>

(22 +24 + x) / 16 > (30 + 42 + 21) /24

(22 +24 + x) > 3.875 (16)

X > 16

<span>So they should collect more than 16 food item to have a greater item per student</span>

5 0

3 years ago

Simplify 20+4(3)÷(-8)

Mashcka [7]

The answer to this problem would be 12. If you need to show work just comment.

4 0

2 years ago

Read 2 more answers

In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo

larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine $z_{\frac{\alpha}{2}}=z_{0.005}$

Now by using z score table we find that $z_{\frac{\alpha}{2}}=2.58$

The boundaries of the confidence interval are:

$\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64$

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

3 0

3 years ago