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3 years ago

Jean needs 1 third cup of walnuts for each serving of salad she makes she has 2 cups of walnuts how many serving can she make

Mathematics

1 answer:

olchik [2.2K]3 years ago

4 0

In 1/3 cup of walnuts, she makes 1 serving of salad

Therefore, In 1 cup of walnuts, she makes 3 servings of sald.

And hence in 2 cups of walnuts, she makes 2*3 = 6 serving of salad.

So the answer of the given question is , if she has 2 cups of walnus, then she makes 6 serving of salad .

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What is 3/7 plus 8/6??

castortr0y [4]

Find a common denominator, then add and simplify

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2 years ago

Can someone help me with this problem?

kompoz [17]

Answer: 5:6

Explanation:

The 2 polygons are similar

So 25/30 = 25.5/30.6 = 5/6 or 5:6

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2 years 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

The function f(x) = (x − 4)(x − 2) is shown.What is the range of the function?

yuradex [85]

Answer:

The range is y ≥ -1

Step-by-step explanation:

∵ f(x) = (x - 4)(x - 2)

∴ f(x) = x² - 2x - 4x + 8

∴ f(x) = x² - 6x + 8 ⇒ quadratic function (ax² + bx + c) represented by

parabola graphically

∵ a = 1 , b = -6 , c = 8

∴ x-coordinate of its vertex = -b/2a = -(-6)/2×1 = 3

∴ f(3) = (3)² - 6(3) + 8 = -1

∵ a is positive ⇒ the curve has minimum point and it's open upward

∴ the minimum point is (3 , -1)

∴ The range is y ≥ -1 ⇒ because the minimum value is -1

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3 years ago

Find x and round to nearest tenth

Lesechka [4]

Answer:

x=13.5

Step-by-step explanation:

Using the formula:

Sin theta=opposite/hypotenuse

Where theta=57 degrees

opposite =10.8

hypotenuse =x

Sin57=10.8/x

Xsin57=10.8

x=10.8/sin57

x=10.8/0.8

x=13.5

8 0

2 years ago