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4 years ago

What is the value of x?

Mathematics

1 answer:

malfutka [58]4 years ago

7 0

Answer:there is no Val to x except I you use it as an variable

Step-by-step explanation:

Do the same steps that you would use for any other problem

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Mrs. Byrne washed

AleksAgata [21]

Answer:

Mrs. Byrne washed more

Step-by-step explanation:

5 0

2 years ago

Please explain how you got this answer.

Tema [17]

150 times

out of the first 5 cards, there is a 3/5 chance of drawing a yellow.
if you replace the card and draw again, you still have a 3/5 chance of drawing a yellow card.
keep doing this process for 250 draws, each draw will always have 3/5 chance of drawing a yellow card.

so 3/5 +3/5 +3/5 +... (250 times) = 250 x 3/5 = 150

3 0

4 years ago

The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {4

Mrac [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

$Mean = x1 = \frac{20+24+40+63+76+89}{6}$

$x1 = \frac{312}{6}$

$x1 = 52$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

$20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37$

The deviations will be added then. $Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24$

So the mean absolute deviation for city 1 is 24 ..

For city 2:

$Mean = x2 = \frac{41+50+58+62+72+83}{6}$

$x2 = \frac{366}{6}$

$x2 = 61$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

$41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22$

The deviations will be added then. $Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33$

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

3 0

3 years ago

Read 2 more answers

PLEASE HELP I WILL GIVE BRAINLIEST

aliina [53]

Answer:

The quadratic equations and their solutions are;

9 ± √33 /4 = 2x² - 9x + 6.

4 ± √6 /2 = 2x² - 8x + 5.

9 ± √89 /4 = 2x² - 9x - 1.

4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

2x² - 8x + 5, here a = 2, b = -8, c = 5. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4. 24 can also be written as 4 × 6 and √4 = 2. So x = 8 ± 2√6 / 2×2= 4±√6/2.

2x² - 10x + 3, here a = 2, b = -10, c = 3. x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.

2x² - 8x - 3, here a = 2, b = -8, c = -3. x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4. 88 can also be written as 4 × 22 and √4 = 2. So x = 8 ± 2√22 / 2×2 = 4± √22/2.

2x² - 9x - 1, here a = 2, b = -9, c = -1. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4. x = 9 ± √89 / 4.

2x² - 9x + 6, here a = 2, b = -9, c = 6. x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4. x = 9 ± √33 / 4 .

Step-by-step explanation:

8 0

3 years ago

Why is it important that the definition of logarithms states that the base of the logarithm does not equal 1?

lys-0071 [83]

The change of base rule says
$\log_b(x) = \frac{\log(x)}{\log(b)}$

If b = 1, then we will have

$\log_b(x) = \frac{\log(x)}{\log(b)}$

$\log_1(x) = \frac{\log(x)}{\log(1)} = \frac{\log(x)}{0}$

which is NOT possible. We cannot divide by zero. So this is why b = 1 is NOT allowed.

5 0

4 years ago

Read 2 more answers