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VashaNatasha [74]
2 years ago
10

Solve the inequality and enter the solution as an inequality X-7<1​

Mathematics
1 answer:
enot [183]2 years ago
5 0

Answer: x < 8

Step-by-step explanation:

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Find x<br> options;<br> a) 21√2/2<br> b) 21√2<br> c) 21√3/2<br> d) 7
pishuonlain [190]

The value of x is  21√2/2

The correct option is (A)

<h3>What is Trigonometry?</h3>

Trigonometry is a branch of mathematics that studies relationships between the sides and angles of triangles.

In triangle having angle 60.

sin 60 = P/ 7√3

√3/2= P/7√3

2P= 21

P= 21/2

Now again

cos 45= 21/2/x

1/√2 = 10.5/x

x=  21√2/2

Learn more about trigonometry here:

brainly.com/question/26719838

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6 0
1 year ago
ILL GIVE BRAINLIEST!! PLS HELP!!! Ten students are taking both algebra and drafting. There are 24 students taking algebra. There
Minchanka [31]

Answer:

15 students

Step-by-step explanation:

from the algebra = 24 - 10 = 14

from the draft = 11 - 10 = 1

14 + 1 = 15

the total students that are taking algebra or drafting but not both is 15 students

4 0
3 years ago
Read 2 more answers
How many kilometers in the world?
Nina [5.8K]
40,000 kilometers in the world
6 0
2 years ago
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D’Quan’s grandmother made a quilt for his bed. The quilt is
damaskus [11]
One Square Meter= 10.764 Square Feet.
The total area of the quilt in meters is 4.465m^{2}
As you're looking for square feet, you've then got to multiply this by 10.764
4.465*10.764= 48.06086 square feet.
Rounded, this is 48.06 square feet.
Hope this helps :) 
7 0
3 years ago
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An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so
alexgriva [62]

Solution :

Let p_1 and p_2  represents the proportions of the seeds which germinate among the seeds planted in the soil containing 3\% and 5\% mushroom compost by weight respectively.

To test the null hypothesis H_0: p_1=p_2 against the alternate hypothesis  H_1:p_1 \neq p_2 .

Let \hat p_1, \hat p_2 denotes the respective sample proportions and the n_1, n_2 represents the sample size respectively.

$\hat p_1 = \frac{74}{155} = 0.477419

n_1=155

$p_2=\frac{86}{155}=0.554839

n_2=155

The test statistic can be written as :

$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}

which under H_0  follows the standard normal distribution.

We reject H_0 at 0.05 level of significance, if the P-value or if |z_{obs}|>Z_{0.025}

Now, the value of the test statistics = -1.368928

The critical value = \pm 1.959964

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

                                     = 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$, so we fail to reject H_0 at 0.05 level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the \text{proportion} of the seeds that \text{germinate differs} with the percent of the \text{mushroom compost} in the soil.

8 0
3 years ago
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