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AleksandrR [38]
3 years ago
6

Please help How many 1/5 cups are in 2/15 cups of cookie dough?

Mathematics
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

There are 2/3 of 1/5 cups in 2/15 cups of cookie dough.

Step-by-step explanation:

1/5=3/15

Divide 2/15 by 3/15 using keep, change, flip rules

2/15x15/3=30/45=6/9=2/3

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12.5% of $100 is what number? percent proption
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Answer: $12.5

Step-by-step explanation:

The $100 Is 100 percent meaning that one percent is $1. Take that and multiply it by 12.5 and get $12.5

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Express each of the following radios in the simplest form.
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Step-by-step explanation:

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2 years ago
Rob and Mike are selling cookie dough
SpyIntel [72]

The cost of one package of white chocolate chip cookie is $9 while the cost of one package of oatmeal cookie dough is $14.

<h3>What is an equation?</h3>

An equation is an expression used to show the relationship between two or more numbers and variables.

Let x represent the cost of white chocolate chip and y represent the cost of oatmeal dough, hence:

12x + 6y = 192   (1)

Also:

6x + 12y = 222    (2)

From both equations:

x = 9, y = 14

The cost of one package of white chocolate chip cookie is $9 while the cost of one package of oatmeal cookie dough is $14.

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1 year ago
If MrBeast bought 69 Lamborghini's and bought 420 Rolex watches how many items in all would he have bought altogether?
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6 0
3 years ago
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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
Oksanka [162]

Answer:

a) [-0.134,0.034]

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

s_p^2 = \frac{ (n_1 -1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2},

where n_1 = 40, n_2 = 30, s_1 = 0.16, s_2 = 0.19.

The mean difference \mu_1 - \mu_2 = 10.85 - 10.90 = -0.05.

The confidence interval is

(\mu_1 - \mu_2) \pm t_{n_1+n_2-2,\alpha/2} \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} = (-0.05) \pm t_{68,0.025} \sqrt{\frac{0.03}{40} + \frac{0.03}{30}}

= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]

b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.

c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.

6 0
3 years ago
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