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3 years ago

Help me with this please

Mathematics

1 answer:

kumpel [21]3 years ago

5 0

Answer:

10w ≥ 30

Step-by-step explanation:

First, the variable w represents the "sum of a number and 15", so we don't know the missing value. Therefore the first part is 10w. The inequality is ≥ because it represents "at least 30". Your inequality should be 10w ≥ 30

Hope this helps!

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after 8% discount from the list price, the sale price price of a poster was $17.48. what was the list price of poster.

Ludmilka [50]

Well the answer would be $18.8784 but if you want it to be shorter it would be $18.88 because you multiply 17.48 and 0.08 and get 1.3984 and then you add it to $17.48

6 0

3 years ago

Which is greater 2.16 or 2.061

jenyasd209 [6]

2.16 is greater. To do this start from the left and compare the numbers. For example, 2 is the same in both but 1 is greater than 0, so the answer is 2.16.

6 0

3 years ago

Read 2 more answers

If (-1, y) lies on the graph of y=2^2x then y=

zubka84 [21]

Answer:

y is 1/4 when x is -1 if I did interpret your equation right. Please see that I did.

I believe the equation to be $y=2^{2x}$ . If you meant $y=2^2x$ , please let me know.

Thanks kindly.

Step-by-step explanation:

If (-1,y) lies on the graph of $y=2^{2x}$ , then by substitution we have:

$y=2^{2\cdot -1}$ .

We just need to simplify to determine y:

$y=2^{2 \cdot -1}$

Multiply the 2 and -1:

$y=2^{-2}$

Use reciprocal rule for exponents to get rid of the negative:

$y=\frac{1}{2^2}$

$2^2$ just means $2(2)$ :

$y=\frac{1}{2(2)}$

$y=\frac{1}{4}$ .

8 0

3 years ago

Past experience has indicated that the breaking strength of yarn used in manufacturing drapery material is normally distributed

vlabodo [156]

Answer:

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm)

Step-by-step explanation:

Our sample size is 11.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 11-1 = 10$ .

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 10 and 0.025 in the two-sided t-distribution table, we have $T = 1.812$

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{4}{\sqrt{11}} = 1.2060$

Now, we multiply T and s

$M = Ts = 1.812*1.2060 = 2.19/tex]For the lower end of the interval, we subtract the sample mean by M. So the lower end of the interval here is[tex]L = 97 - 2.19 = 94.81 = 94.8$ cm

For the upper end of the interval, we add the sample mean and M. So the upper end of the interval here is

$L = 97 + 2.19 = 99.19 = 99.2$ cm

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm).

7 0

3 years ago

Assuming x > 0, which of these expressions is equivalent to 11 times the square root of 245 x to the third plus 9 times the s

ehidna [41]

The answer is <span>104 x times the square root of 5 x
</span>
<span>11 times the square root of 245 x to the third plus 9 times the square root of 45 x to the third is:
</span> $11 \sqrt{245 x^{3} } +9 \sqrt{45 x^{3} } =11 \sqrt{245} *\sqrt{ x^{3} } +9 \sqrt{45} *\sqrt{ x^{3} }= \\ \\ =11 \sqrt{5*49} * \sqrt{ x^{2} *x} +9 \sqrt{5*9} * \sqrt{ x^{2} *x}= \\ \\ =11* \sqrt{5} * \sqrt{49} * \sqrt{ x^{2}} * \sqrt{x} +9* \sqrt{5} * \sqrt{9} * \sqrt{ x^{2} }* \sqrt{x} = \\ \\ 
=11* \sqrt{5} *7*x* \sqrt{x} +9* \sqrt{5} *3*x* \sqrt{x} = \\ \\ 
=77x* \sqrt{5*x} +27x* \sqrt{5*x} = \\ \\ 
=77x \sqrt{5x} +27x \sqrt{5x} = \\ \\ 
=104x \sqrt{5x}$

4 0

3 years ago

Read 2 more answers

Help me with this please ​

Help me with this please