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Len [333]
2 years ago
9

A piece of wire is 7/8m long broke into two piece one piece is 1/4 m long how long is the other one ?​

Mathematics
1 answer:
hjlf2 years ago
7 0

Answer: 5/8m

Step-by-step explanation:

From the question, we are informed that a piece of wire is 7/8m long and broken into two piece. We are further told that one piece is 1/4 m long.

The length of the other piece of wire would be calculated as:

= 7/8 - 1/4

= 7/8 - 2/8

= 5/8m

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AnnyKZ [126]
53 is the answer, I think
4 0
3 years ago
Please help me........​
scZoUnD [109]

Answer:

a)5.8meters

b)15feet

c)She is right as 16 feet is equivalent to 4.8768m which is less than 5m

d)3.6576m

Step-by-step explanation:

7 0
2 years ago
Solve the following inequality. |3n-2|-2<1
disa [49]

Answer: \frac{-1}{3} < n < \frac{5}{3}

<u>Step-by-step explanation:</u>

| 3n - 2 | - 2 < 1

<u>               +2 </u> <u>+2 </u>

| 3n - 2 |       < 3

3n - 2 < 3     and     3n - 2 > -3

<u>      +2 </u> <u>+2   </u>              <u>     +2 </u>  <u> +2 </u>

3n       < 5      and     3n      > -1

       n <  \frac{5}{3}       and         n > \frac{-1}{3}

\frac{-1}{3} < n < \frac{5}{3}

Interval Notation:  (\frac{-1}{3},\frac{5}{3})

Graph:  \frac{-1}{3} o--------------------o \frac{5}{3}


7 0
3 years ago
Read 2 more answers
Which number line below correctly represents the expression -3 + 4?
Archy [21]

Answer:

1st answer

Step-by-step explanation:

First you subrtract 3 from 0 then add 4 to -3

8 0
3 years ago
National data indicates that​ 35% of households own a desktop computer. In a random sample of 570​ households, 40% owned a deskt
Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

Step-by-step explanation:

We are given that National data indicates that​ 35% of households own a desktop computer.

In a random sample of 570​ households, 40% owned a desktop computer.

<em><u>Let p = population proportion of households who own a desktop computer</u></em>

SO, Null Hypothesis, H_0 : p = 25%   {means that 35% of households own a desktop computer}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 570​ households who owned a desktop computer = 40%

            n = sample of households = 570

So, <u><em>test statistics</em></u>  =  \frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }

                               =  2.437

<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

3 0
2 years ago
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