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NikAS [45]
2 years ago
12

Write equivalent fractions for 2 4/5 and 2 2/3 using a common denominator.

Mathematics
2 answers:
Stolb23 [73]2 years ago
6 0

Answer:

2 4/5=2 12/15

2 2/3=2 10/15

both with a common denominator of 15

hope this helps

have  a good day :)

Step-by-step explanation:

Luda [366]2 years ago
5 0

Answer:

2 12/15 and 2 10/15

Step-by-step explanation:

4 x 3 = 12

2 x 5 = 10

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Solve for x.
White raven [17]

Step-by-step explanation:

25(x - 910) =  - 234 \\ 25x - 22750 =  - 234 \\ 25x = 22516 \\ x =  \frac{22516}{25}  \\ x = 900 \frac{16}{25}

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2 years ago
Help solve 88 please
Fofino [41]

Answer:

The equation of the line is y = -2x + 7

Step-by-step explanation:

* Lets study how to make an equation of the line from two point

- If the line passes through points (x1 , y1) and (x2 , y2)

- The form of the equation is y = mx + c, where m is the slope of the

 line and c is the y-intercept

- The rule of the slope is m = (y2 - y1)/(x2 - x1)

- The y-intercept means the line intersect the y-axis at point (0 ,c)

* Now lets solve the problem

∵ The line passes through points (-3 , 13) and (4 , -1)

- Let (-3 , 13) is (x1 , y1) and (4 , -1) is (x2 , y2)

∵ m = (y2 - y1)/(x2 - x1)

∴ m = (-1 - 13)/(4 - -3)

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- Lets write the form of the equation

∵ y = mx + c ⇒ substitute the value of m

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4 0
2 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.

7 0
3 years ago
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