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4 years ago

What one is bigger a liter of ketchup or 750 milliliter of ketchup

Mathematics

2 answers:

djverab [1.8K]4 years ago

5 0

Well a l is 1000ml, u do the math

erik [133]4 years ago

4 0

1 liter of ketchup hope it helps

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BRAINLIEST ********

timama [110]

Answer:

$ 1249.6

Step-by-step explanation:

given,

cost of saddle = $534.00

riding chaps cost = $87.02

boots cost = $143.99

hat cost = $55.87

pairs of glove cost = $15.94

sales tax paid = 7.5%

Sam's lesson cost = $35.00 per lesson

number of lesson = 10

Total purchase cost

= $534.00 + $87.02 + $143.99 + $55.87 + $15.94

= $ 836.82

cost included sales tax

= 1.075 x $ 836.82

= $ 899.58

Cost of lesson = 10 x $35.00 = $350.0

Total cost of the Sam

= $ 899.58 + $350.0

= $ 1249.58

= $ 1249.6

7 0

4 years ago

What is two decimal that are equivalent to 7.06

ycow [4]

Answer:7.060 and 7.0600

Step-by-step explanation:0 doest not mean nothing

4 0

3 years ago

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Fill in the missing angle measure

kykrilka [37]

Step-by-step explanation:

please mark me as brainlest

4 0

3 years ago

Read 2 more answers

Polar form of (x+6)^2 +y^2=36

Sindrei [870]

Answer:

r = -12cos(θ)

Step-by-step explanation:

The usual translation can be used:

x = r·cos(θ)
y = r·sin(θ)

Putting these relationships into the formula, we have ...

(r·cos(θ) +6)² +(r·sin(θ))² = 36

r²·cos(θ)² +12r·cos(θ) +36 +r²·sin(θ)² = 36

r² +12r·cos(θ) = 0 . . . . subtract 36, use the trig identity cos²+sin²=1

r(r +12cos(θ)) = 0

This has two solutions for r:

r = 0 . . . . . . . . a point at the origin

r = -12cos(θ) . . . the circle of interest

4 0

4 years ago

Find the standard deviation. Round to the nearest tenth. 1, 2, 11, 8, 16, 16,20, 16, 18 OA. 6.5 O B. 6.9 O C. 1.5 ? D. 7.4

NikAS [45]

Answer:

A. 6.5

Step-by-step explanation:

First we find the average $\bar{x}$ of the 9 data:

$\bar{x} =\frac{\sum_{x=1}^{n}x_{i}}{n}$

Where n is the data number, that in this case is 9.

$\bar{x} =\frac{1+ 2+ 11+ 8+ 16+ 16+20+ 16+ 18}{9}=12\\$

The formula of the standard deviation $\sigma$ is:

$\sigma=\sqrt{\frac{\sum_{x=1}^{n}(x_{i}-\bar{x})^{2}}{n}}$

We replace the data and find the value of the standard deviation:

$\sigma=\sqrt{\frac{(1-12)^{2}+(2-12)^{2}+(11-12)^{2}+(8-12)^{2}+(16-12)^{2}+(16-12)^{2}+(20-12)^{2}+(16-12)^{2}+(18-12)^{2}}{9}}$

$\sigma=\sqrt{\frac{(-11)^{2}+(-10)^{2}+(-1)^{2}+(-4)^{2}+(4)^{2}+(4)^{2}+(8)^{2}+(4)^{2}+(6)^{2}}{9}}=6,54$

We approximate the number and the solution is 6,5

6 0

4 years ago