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Alex777 [14]
2 years ago
10

Write an expression with the same meaning as "7 less than the square of a number."

Mathematics
1 answer:
Artemon [7]2 years ago
7 0

Answer:

x^2-7 is the correct answer

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I need help please .
levacccp [35]

Answer:

Step-by-step explanation:

You know that anything to the zeroeth power is 1, so the equation is now

-7 * 1, or -7.

7 0
2 years ago
100^m ÷ 1000^n can be written in the form 10^z. Express z in terms of m and n
Kitty [74]

100m ÷ 1000n

(102)m ÷ (103)n

102m ÷ 103m

10-m

4 0
3 years ago
Read 2 more answers
What are the possible rational zeros of f(x) = x4 − 4x3 + 9x2 + 5x + 14?
Dovator [93]

Answer:

±1, ±2, ±7, ±14

Step-by-step explanation:

6 0
3 years ago
49/10 as a mixed number
anastassius [24]
4 9/10 is your answer because you first have to divide the numerator by the denominator. this will give you 4, with a remainder of 9. The mix number can be created by using the 4 as the whole number, the 9 as the numerator and the 10 as a denominator. So, 49/10 = 4 9/10. Hope this works!
5 0
3 years ago
Read 2 more answers
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
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