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rewona [7]
2 years ago
8

Mr Torres made punch for the Valentine's day party the recipe called for 2 scoops of ice cream for every 1 1/2 cup of soda. if M

r Torres added 20 scoops of ice cream but only had a 1/4 cup measuring cup how many times will he have to fill it in order to make the punch?​
Mathematics
1 answer:
Katen [24]2 years ago
3 0

Answer:

20 i think dont answer yet

Step-by-step explanation:

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Help me please ThankYou !!
matrenka [14]
My answer would be A.
3 0
3 years ago
Can anyone please help me with this question
Ede4ka [16]

9514 1404 393

Answer:

  y ≥ 1/4x + 2

Step-by-step explanation:

The line is solid, and the shading is above it, so the inequality will have a form similar to ...

  y ≥ [the boundary line]

The y-intercept is y = 2, and the slope is 1 unit up for 4 units right, so 1/4. In slope-intercept form, the equation of the boundary line is ...

  y = mx + b . . . . . . for slope m and y-intercept b

This boundary line has equation ...

  y = 1/4x + 2

So, the inequality is ...

  y ≥ 1/4x + 2

4 0
3 years ago
3.95174 rounded to nearest hundredths
OlgaM077 [116]

The hundredths place is where the 5 is.

Look to the right of the hundredths place and determine if the number is 5 or more or if the number is 4 or less.

In this problem, the number to the right of 5 is 1, and 1 is less than 4.

This means you can keep the 5 the same and every number after the 5 becomes imaginary zeroes.

3.95174 becomes 3.95 (rounded to nearest hundredth).

4 0
3 years ago
Read 2 more answers
The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of sixte
goldenfox [79]

Answer:

Yes, it can be concluded that the mean hexane concentration is less in treated water than in unsaturated water

Step-by-step explanation:

The number of of specimen in the samples of untreated water, n₁ = 16

The sample mean, \overline x_1 = 228.0

The sample standard deviation, s₁ = 4.3

The number of of specimen in the samples of treated water, n₂ = 20

The sample mean, \overline x_2 = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; \overline x_1 ≥ \overline x_2

The alternative hypothesis, Hₐ;  \overline x_1 < \overline x_2

The degrees of freedom = 16 - 1 = 15

The test statistic, t_{\alpha} = 1.341

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

Plugging in the values, we get;

t=\dfrac{(224.6- 228.0)}{\sqrt{\dfrac{5.0^{2}}{20} -\dfrac{4.3^{2} }{16}}} \approx -11.0675

Given that the t-value is large, the corresponding p-value is low, therefore, we fail to reject the null hypothesis and there is considerable statistical evidence to suggest that the mean hexane concentration is less in treated than in untreated water, therefore, we have; \overline x_1 ≥ \overline x_2

6 0
3 years ago
If UX=34 and VX=37, what is WX? Write your answer as a whole number or as a decimal rounded to the nearest hundredth. *
PolarNik [594]
I believe the answer would be WX=39
8 0
3 years ago
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