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2 years ago

Six more than five times a number is at least twenty-one

1 answer:

6 0

Lets say the number is x so we can make an equation
If is at least 21, that means it has to be greater then or equal to it.
6+5*x>=21
You can subtract 6 from both sides
5*x>=15
Then divide both sides by 5
x>=3
You can conclude that the number is greater then or equal to 3.

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Starting at sea level, a submarine descended at a constant rate to a depth of − 4/5 mile relative to sea level in 3 minutes.

Jobisdone [24]

If it were 4 then it would be 1 1/3 for one minute

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2 years ago

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Lupe bought 8 pounds of pretzels at a local wholesaler for $18. Her friend Charles bought 6 pounds of pretzels at the supermarke

Novay_Z [31]

Answer:

Step-by-step explanation:

Charle would have spent that much because one pound is 2.50

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2.50x8=20

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2 years ago

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Help!!!!!!!!!!!!!!!!!!!!plz

motikmotik

Descriptions 1, 2, 3, and 5 are accurate, though 5 is ambiguous.

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Description 4 goes with (7x)² = 49x², not 7x².

Description 6 goes with 5·8 + x² + (2x + 8) = x² +2x +48, not 40x² +100x +40.

The description for 5 is accurate, but could also be applied to y² + 3(y - 4), which is why it can be considered to be ambiguous. There are a number of ways to resolve the ambiguity, but they were not used here.

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2 years ago

Estimate a 15% tip on a dinner bill of $52.87 by first rounding the bill amount to the nearest ten dollars.

Stels [109]

Answer:

$ 7.9

Step-by-step explanation:

Amount of bill = $52.87 = $52.9

Amount of tip = 15% of $52.9

$= \frac{15}{100} \times 52.9 \\ \\ = 0.15 \times 52.9 \\ = 7.935 \\ = \$ \: 7.9 \\$

8 0

2 years ago

The marketing director of a large department store wants to estimate the average number of customers who enter the store every f

Dahasolnce [82]

Answer:

$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and $t_{\alpha /2}$ by 2.2621 we get:

$49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326$

Finally, the interval values that she get is:

$36.5674\leq x'\leq61.4326$

8 0

2 years ago