All categories

3 years ago

15 crackers weigh 69 grams. how many kilograms is this?

Mathematics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

$0.069\ kg$

Step-by-step explanation:

we know that

1 kg=1,000 g

using proportion

Find out how many kilograms are 69 grams

Let

x -----> the weight in kg

$\frac{1}{1,000}\frac{kg}{g} =\frac{x}{69}\frac{kg}{g}\\\\x=\frac{69}{1,000}\\\\x=0.069\ kg$

You might be interested in

a tree casts a shadow 32 feet long .the Angle of elevation of the sun is 62° what is the height of the tree

Delicious77 [7]

$\tan \theta = \dfrac{opp}{adj}$

$\tan 62^\circ = \dfrac{y}{32~ft}$

$y = 32 \tan 62^\circ ~ft$

$y = 60.2 ~ft$

Answer: The tree is 60.2 ft tall.

4 0

3 years ago

Read 2 more answers

For what value of x is line m parallel to line n?

kogti [31]

Answer:

$x=10$

Step-by-step explanation:

Here, we can use an angle rule.

We see that angles $(8x+50)$ ° and 130° are under a sort of F shape, where lines m and n are the horizontal parts of the letter and where the diagonal line is the vertical part of the letter.

This means we can use the rule:

Corresponding angles are equal.

This means that:

$8x + 50 = 130$

We can now use this equation we have formed to solve for $x$ :

$8x + 50 = 130\\8x = 80\\x = 10$

Therefore, our final answer is $x=10$ .

3 0

1 year ago

Question 9 (1 point)

Romashka [77]

Answer:

Step-by-step explanation:

League A League B

151.12 163.25

148 157

26.83 24.93

29 136

136 145

167 178

207 256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(82.5-1.5\times 76.56,159.06+1.5\times 76.56)$

(-32.34,273.9)

So, There is no outlier

Maximum = 207

League B in ascending order :

24.93,136,145,157,163.25,178,256

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

$Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5$

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(140.5-1.5\times 30.125,170.625+1.5\times 30.125)$

(95.3125,215.8125)

24.93 and 256 are outliers

Maximum = 256

5 0

3 years ago

At a pizza shop, you can choose thick or thin crust, red or white sauce, and toppings of pepperoni, cheese, and onion. How many

lozanna [386]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

What number should go in box A on this number line? 4. 5. A

cestrela7 [59]

Answer:

Step-by-step explanation:

4 0

2 years ago