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shusha [124]
3 years ago
9

Omar was born on December 1, 1998, Omar's mom was born on January 18, 1965.

Mathematics
1 answer:
worty [1.4K]3 years ago
3 0

Answer:

33 yrs old

Step-by-step explanation:count up to the year he was born

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Solve the inequality.<br><br> 5/6t - 3≥3t + 6
Papessa [141]

Answer:

t ≤ -\frac{54}{13}

Step-by-step explanation:

\frac{5}{6}t−3−3t≥3t+6−3t

-\frac{13}{6}t−3≥6

-\frac{13}{6}t−3+3≥6+3

\frac{13}{6}t≥9

-\frac{6}{13}(-\frac{13}{6}t) ≥ -\frac{6}{13}(9)

t ≤ -\frac{54}{13}

6 0
2 years ago
9. Determine if x + 1 is a factor of x ^ 4 - 4x ^ 2 - 4x - 1 Justify your answer.
Elden [556K]

Answer:

No

Step-by-step explanation:

x + 1 is not a factor of the polynomial because the remainder is 2 and not 0

5 0
2 years ago
Expanded form: (4a+4b)^3
TiliK225 [7]

Step-by-step explanation:

use the formula: (a+b)³ = a³+3a²b+3ab²+b³

and place the corresponding values.

7 0
3 years ago
Rearrange the equation so xxx is the independent variable. 6x+y=4x+11y6x+y=4x+11y
slava [35]

Answer:

10x=10y so x and y are the same or x=y

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances
wariber [46]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.

So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.

Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:

X₁: Braking distance of an SUV equipped with tires made with compound one.

Its sample mean is X[bar]₁= 69 feet

And the Standard deviation S₁= 10.4 feet

X₂: Braking distance of an SUV equipped with tires made with compound two.

Its sample mean is X[bar]₂= 71 feet

And the Standard deviation S₂= 7.6 feet

We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.

The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

The statistic to use to compare these two populations is a pooled Z test

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } }

Z ≈ N(0;1)

Z_{H_0}= \frac{69-71-0}{\sqrt{\frac{108.16}{81} +\frac{57.76}{81} } }= -1.397

The rejection region if this hypothesis test is one-tailed to the right, so you'll reject the null hypothesis to small values of the statistic. The critical value for this test is:

Z_{\alpha  } = Z_{0.05}= -1.648

Decision rule:

If Z_{H_0} > -1.648 , then you do not reject the null hypothesis.

If Z_{H_0} ≤ -1.648 , then you reject the null hypothesis.

Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.

At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.

I hope you have a SUPER day!

4 0
3 years ago
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