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loris [4]
3 years ago
15

Helpppppppp meeeeeeeeeeee

Mathematics
1 answer:
sertanlavr [38]3 years ago
3 0
A and E.

x= (1+ root-71)/6 and x= (1-root-71)/6
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There are two factors of -36 such that one factor is 11 less than half of the other factor choose all the pairs of these factors
Amanda [17]

Answer with explanation:

Factors of  -36 are

      =\pm 1, \pm 2,\pm 3,\pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36

It is given that , among the factors of , -36 , one factor is 11 less than half of the other factor.

If you will take,  =\pm 1,\pm 3,\pm 4, \pm 9,, then you can find such pairs.

Pairs are= (3,-4), (9, -1),(-1, -6),

7 0
2 years ago
Juan has 12 pencils. two-thirds of these pencils are sharpned which model represents the number of sharpend pencils juan has?
Oduvanchick [21]

Answer:

8

Step-by-step explanation:

8 0
2 years ago
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Two batches of soup will serve 8 people. Lonnie needs to make enough soup for 120 people. If the recipe calls for 2 onions for o
Inessa05 [86]
<span><em>"2 batches of soup will serve                            8 people."</em>
<u>Then x </u></span><span><u>batches of soup will serve                         120 people         
</u>120.2 = 8.x
x = 30 batches of soup will serve 120 people.

<em>"2 onions for                1 batch"</em>
<u>Then y onions for         30 batches                                                          
</u>30.2 = 1.y
y = 60 onions<u>




</u></span>
5 0
3 years ago
Read 2 more answers
Historically, a certain region has experienced 92 thunder days annually. (A "thunder day" is day on which at least one instance
DiKsa [7]

Answer:

We conclude that the mean number of thunder days is less than 92.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 92 thunder days annually.

Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38.

<u><em>Let </em></u>\mu<u><em> = mean number of thunder days.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 92 days     {means that the mean number of thunder days is more than or equal to 92}

Alternate Hypothesis, H_A : \mu < 92 days     {means that the mean number of thunder days is less than 92}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean number of thunder days = 72

             s = sample standard deviation = 38

            n = sample of years = 15

So, <u><em>test statistics</em></u>  =  \frac{72-92}{\frac{38}{\sqrt{15} } }  ~ t_1_4

                               =  -2.038

The value of z test statistics is -2.038.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t as -2.038 < -1.761, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean number of thunder days is less than 92.

7 0
3 years ago
Dora bought a bottle of nail polish that was marked down by 20 percent from its original price of $4.50. Including a 9 percent s
o-na [289]

Answer:

3.92

Step-by-step explanation:

7 0
3 years ago
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