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4 years ago

Jonas spends an average of $3.42 on meals each day. How much does he spend on meals each week?

Mathematics

2 answers:

LenaWriter [7]4 years ago

4 0

He Spends $24.64 Every Week
Hope This Helps

MArishka [77]4 years ago

3 0

I think it is 41.04...........................

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Level 1-2 1. Sally picked 5 apples, but she needs 12 to bake 3 apple pies. How many more apples does Sally need? (To access the

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Step-by-step explanation:

sally needs 7 more Apples, but if Sally is doubling the recipe she will need 14 more Apples

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3 years ago

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kogti [31]

Answer:

8.75 per hour

Step-by-step explanation:

Since it is a linear function we can use the slope formula

m = ( y2-y1)/(x2-x1)

= ( 370-20)/(40-0)

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3 years ago

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4 years ago

Scores on a national math test are Normally distributed, with a mean score of 490 and a standard deviation of 50. Which test sco

alisha [4.7K]

Answer:

532

Step-by-step explanation:

The score that represents the top 20% = A score = 80%

Z score for 80% percentile = 0.842

We solve using the z score formula

z = (x-μ)/σ, where

x is the raw score = ??

μ is the population mean = 490

and σ is the population standard deviation = 50

Hence:

0.842 = x - 490/50

Cross Multiply

0.842 × 50 = x - 490

x = 0.842 × 50 + 490

x = 42.1 + 490

x = 532.1

Approximately = 532

Therefore, the test score = 532

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3 years ago

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Find the value of the variable y, where the sum of the fraction 2/y-3 and 6/y+3 is equal to the quotient.

NISA [10]

Answer:

Here we need to solve:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }$

The sum of the fractions is equal to the quotient between the fractions.

Notice that the two values:

y = 3

y = -3

make the denominator equal to zero, so those values are restricted.

We can simplify the right side to get:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)} = 3*\frac{y + 3}{y - 3}$

Now we can multiply both sides by (y - 3)

$(y - 3)*(\frac{2}{y - 3} + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)$

Now we can multiply both sides by (y + 3)

$(2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)$

$2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45$

First, let's see the determinant of that quadratic equation:

$D = (-2)^2 - 4*3*45 = -536$

We can see that it is negative, thus, there are no real solutions of the equation.

Thus, there is no value of y such that the origina equation is true,

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3 years ago