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Mariana [72]
3 years ago
13

18 ft = 6 yd O True O False Help

Mathematics
2 answers:
Tomtit [17]3 years ago
8 0

Step-by-step explanation:

there is 3 feet in a yard

divide by 3

18 ÷ 3 = 6

true

a_sh-v [17]3 years ago
3 0
It is true. because 18 ft equals 6 yards. hahaha ima just type because it needs to be 20 characters but yeah.
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Need help with my homework 2 problems
Solnce55 [7]
4(x+2)  ,   2(2x+4)  ,  8+4x  ,   1/2(8x+16) are all the same as 4x+ 8.
5 0
3 years ago
5z^3 - 2z^4 - 9z^2 + z
likoan [24]

Answer:

-z( 2z^3-5z^2+9z-1)

Step-by-step explanation:

Apply exponent rule: a^b+c=a^b*a^c

z^2=zz

z^3=z^2*z

z^4=z^3*z

Factor out common term z:

-z(2z^3-5z^2+9z-1)

8 0
3 years ago
Each ride at the fair costs $2.50. Alex will spend at least $15 for rides. Write an inequality to determine the least number of
goblinko [34]
0<r<6
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4 0
3 years ago
Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}

    =\frac{(55-1)\times 94.7}{75}\\\\=68.184

The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0
3 years ago
23 mins after 9 in time can you help me please
Stolb23 [73]

Answer:

9:23 simple

Step-by-step explanation:

4 0
3 years ago
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