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sleet_krkn [62]
3 years ago
5

Violet is baking a mixed berry pie that contains blueberries, cherries, blackberries, and raspberries. She uses three times as m

any blackberries as cherries, twice as many blueberries as raspberries, and the same number of blackberries and raspberries. If Violet has 10 cherries, how many of each of the other berries must she use
Mathematics
2 answers:
erastova [34]3 years ago
7 0

Answer:

Blackberries= 30

Rasberries=30

Blueberries=60

Step-by-step explanation:

10 x 3 = 30 blackberries

Black berries is the same amount as rasberries

30 x 2= 60 blueberries

Svetlanka [38]3 years ago
6 0
Violet has 10 cherries, 30 blackberries, 30 raspberries and 60 blueberries.
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Find the solutions of the quadratic equation 14x^2+9x+10=014x
Vesnalui [34]

Answer:

Option B. x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

14x^{2}+9x+10=0

so

a=14\\b=9\\c=10

substitute in the formula

x=\frac{-9(+/-)\sqrt{9^{2}-4(14)(10)}} {2(14)}

x=\frac{-9(+/-)\sqrt{-479}} {28}

Remember that

i=\sqrt{-1}

substitute

x=\frac{-9(+/-)\sqrt{479}i} {28}  

x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

5 0
3 years ago
A research study investigated differences between male and female students. Based on the study results, we can assume the popula
garri49 [273]

Using the <u>normal distribution and the central limit theorem</u>, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of \mu = 3.5.
  • The standard deviation is of \sigma = 0.5.
  • Sample of 100, hence n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05

The interval that contains 95.44% of the sample means for male students is <u>between Z = -2 and Z = 2</u>, as the subtraction of their p-values is 0.9544, hence:

Z = -2:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

-2 = \frac{X - 3.5}{0.05}

X - 3.5 = -0.1

X = 3.4

Z = 2:

Z = \frac{X - \mu}{s}

2 = \frac{X - 3.5}{0.05}

X - 3.5 = 0.1

X = 3.6

The interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

You can learn more about the <u>normal distribution and the central limit theorem</u> at brainly.com/question/24663213

7 0
2 years ago
(1,3), m = - 3/4 using that as a fraction write an equation in point-slope form
sukhopar [10]
Y=mx+b   m=-3/4  (1,3)
3=-3/4(1)+b
3=-3/4+b
b=3 3/4

y=-3/4+3 3 3/4
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3 years ago
$1.68 for 8 ounces
Shkiper50 [21]
What do you want me to do with that?
8 0
2 years ago
Read 2 more answers
If x&gt;2, then x^2-a-6 / x^2-4=
nadezda [96]
We know that x > 2 ( or : x ≠ +/- 2 )
We have to factorize the numerator and the denominator:
x² - x - 6 = x² - 3 x + 2 x - 6 = x ( x - 3 ) + 2 ( x - 3 ) = ( x - 3 ) ( x + 2 )
x ² - 4 = ( x - 2 ) ( x + 2 )
\frac{(x-3)(x+2)}{(x-2)(x+2)}= \frac{x-3}{x-2}


7 0
3 years ago
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