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tigry1 [53]
3 years ago
5

One bucket of gravel has a mass of 7.05 KG. What is the mass of 20 buckets of gravel in kilograms?

Mathematics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

D: 141 KG

Step-by-step explanation:

7.05 x 20 = 141

Just multiply the mass of the object by however many objects there are if they all weigh the same.

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5/2x+1/4=12 solve for x
aniked [119]

Answer:

4/5 (2x + 5) − 4 =1

4/5(2x +5) = 1 +4

4/5 (2x +5) = 5

2x + 5 = 5 / (4/5)

2x + 5 = 5 (5/4)

2x + 5 = 25/4

2x = 25/4 - 5

2x = 25/4 - 20/4

2x = 5/4

x = (5/4) / 2

x = (5/4) (1/2)

x = 5/8

x = 0.625

7 0
3 years ago
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Help plzz! 13 points and will give brainliest!
laila [671]

Answer: This is rly confusing

Step-by-step explanation:

5 0
3 years ago
Hey guys does anyone know how to lesson 26 Exit ticket 5.4 eureka math engage ny? The last problem: Larry spends half of his wor
Vesnalui [34]

Answer:

\frac{1}{12}X

Step-by-step explanation:

Let X is the number of hours Larry works a day

Given:  Larry spends half of his workday teaching piano lessons.

=> Number of hours he teaches piano a day is: \frac{1}{2}X

As we know that, he has 6 students, so the fraction of his workday is spent with each student is:

\frac{1}{2}X : 6

= \frac{1}{12}X

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Hayden earns $9.50 per hour as a lifeguard.He earns a bonus of $50.00 for taking a training course.How many hours must he work t
mel-nik [20]

Answer: he must work for at least 26.32 hours

Step-by-step explanation:

Let x represent the number of hours that Hayden works as a life guard.

Hayden earns $9.50 per hour as a lifeguard. This means that the amount that he earns for working for x hours would be

9.5x

He earns a bonus of $50.00 for taking a training course. If he takes this bonus cost, the amount that he would earn for working for x hours is

9.5x + 50

Therefore, the number of hours that he must work to earn at least $300.00 would be

9.5x + 50 ≥ 300

9.5x ≥ 300 - 50

9.5x ≥ 250

x ≥ 250/9.5

x ≥ 26.32

4 0
3 years ago
Many elementary school students in a school district currently have ear infections. A random sample of children in two different
Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school n_1 = 42

sample size for the second school n_2  = 34

so 16 out of 42 i.e x_1 = 16 and 18 out of 34 i.e x_2 = 18 have ear infection.

the proportion of students with ear infection Is as follows:

\hat p_1 = \dfrac{16}{42} = 0.38095

\hat p_2 = \dfrac{18}{34}  =  0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}

\bar p = \dfrac{16 +18}{42 +34}

\bar p = \dfrac{34}{76}

\bar p = 0.447368

\bar p + \bar  q = 1 \\ \\ \bar q = 1 -\bar  p \\  \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632

The pooled standard error can be computed by using the formula:

S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} +  \dfrac{\bar p \bar p}{n_2} }

S.E = \sqrt{ \dfrac{  0.447368 *  0.552632}{ 42} +  \dfrac{ 0.447368 *  0.447368}{34} }

S.E = \sqrt{ \dfrac{  0.2472298726}{ 42} +  \dfrac{ 0.2001381274}{34} }

S.E = \sqrt{ 0.01177284105}

S.E = 0.1085

The test statistics is ;

z = \dfrac{\hat p_1 - \hat p_2}{S.E}

z = \dfrac{0.38095- 0.5294}{0.1085}

z = \dfrac{-0.14845}{0.1085}

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0
3 years ago
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