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il63 [147K]
3 years ago
12

A party mix has 8 ounces of pretzels,3 ounces of mini marshmallows, and 6 ounces of nuts. how many ounces of nuts are there for

every ounce of pretzels?
Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0
Hey! Sorry that no one has answered your question in 4 days. But anyways, to do this, we need to simplify the problem a little. The mini marshmallows are unneeded, so we don't even have to think about them. Do do this, all you need to do is divide the 6 ounces of nuts and the 8 ounces of pretzels, or 6 / 8, to get 0.75.
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part of the river she observed fell 1.68 millimeters per year for 1.5 years. Use the distributive property to calculate the exac
Alecsey [184]
1.68 mm per year for 1.5 years could be rewritten as 
(2.00-0.32) mm per year for 1.5 yrs.

Multiply:  (1.5 yrs)(2.00-0.32)(mm/yr) = 3 mm + 0.48 mm = 3.48 mm

This approach makes use of the distributive property of multiplication.
8 0
3 years ago
Simplify
enot [183]

5d(7-3)= 20d

20d-16d= 4d

3x2d= 6d

4d+6d= 10d

the answer you are looking for is 10d

7 0
3 years ago
A bag contains 15 green, 18 yellow, and 16 orange balls. One ball is randomly selected.
sergeinik [125]

Answer:

Green: 15/49

Yellow: 18/49

Orange: 16/49

Step-by-step explanation:

Add the total amt. of balls, then divide a specific group of balls by the total # of balls to get the ratios.

5 0
3 years ago
Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and child
dolphi86 [110]

Answer:

(A) The additional information that is needed to confirm about the conditions for this test have been met is ‘Population is approximately normal.

(B) The test statistic = 1.9965

(C1) P-value = 0.0556

(C2) There is no significant difference between mean VOT for children and adults.

D1) It is possible to make type II error.

(D2) There should be a difference in the VOT of adults and children.

Step-by-step explanation:

Let na be the number of adults = 20

xa mean VOT for adults = 88.17

sa standard deviatiation of VOT for adults = 24.74

Let nc be the number of children = 10

xc mean VOT for children =60.67

sc standard deviatiation of VOT for children = 39.89

(A) From the information the population variances are unknown and the two sample are assumed to be independent and the sample the sample size are smaller that is (n<30).

This indicates that the additional information that is required for the conditions of the test to be satisfied is 'distribution of the population'. the addition assumption to be made is, that the population distribution is normal.

(b) Calculating the test statistics using the formula;

t = (xa -xc)/SE - d

where SE = standard deviation , d= hypothesized difference = 0

But SE = √sa²/na +sc²/nc

           = √24.74 ²/20 + 39.89²/10

           = √189.72559

           = 13.774

Substituting into test statistics equation, we have

t = (xa -xc)/SE - d

  = (88.17 - 60.67/13.774

  = 1.9965

Therefore the test statistic is 1.9965

(c) Calculating the p-value, we have;

Degree of freedom = na + nc -2

                                 = 10+ 20 -2

                                 = 28

The p-value for t=1.9965 at 28 degrees of freedom and 0.05 level of significance is 0.0556.

The p-value 0.0556 is greater than given level of significance 0.05 hence we fail to reject the null hypothesis and conclude that there is no significant difference between mean VOT for children and adults.

(D1) From the information in part (C) the null hypothesis is not rejected.

Since the null hypothesis is not rejected, there might be a chance that not rejecting null hypothesis would be wrong. In this type of situation the error that can occur would be type II error.

(D2) The type of error describe in the context of this study is obtained by the concept of the type II error which tells that the null hypothesis is not rejected when it is actually false.

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3 years ago
2(4t-2)+t=16-t solution set is
ser-zykov [4K]
T=2 the solution set is t=2
5 0
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