A rectangle is twice as long as it is wide. If its area is 50 sq. yd., find the length.

3/4x3-2= -1/4+1/2x+5

Ksju [112]

Answer:

what I think the answer is ×=-9

3 0

3 years ago

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What is the answer to the equation 6x-x/x+1

avanturin [10]

Answer:

5x/x+1

Step-by-step explanation:

Numerator: 6x minus an x equals 5x.

Denominator: Since x and 1 are not like terms, they can't be combined. You just leave them as they were.

Hope this helps!

6 0

3 years ago

Find x and Y pls help

lisabon 2012 [21]

Answer:

Step-by-step explanation:

Comment

The angle supplementary to x and 3x are equal.

Supplementary angles = 180

Solution

x + 3x = 180

4x = 180 Divide by 4

4x/4 = 180/4

x = 45

The angle supplementary to x is 3x

3x = 3*45

3x = 135

This angle is vertically opposite y. Vertically opposite angles are equal. So y = 135

Answer

x = 45

y = 135

3 0

2 years ago

Which of the following is the best estimate of 7/8? ( Answer ASAP )

Triss [41]

Answer:

C

Step-by-step explanation:

just round up to nearest number and 7/8 is 0.87 so it coloser to 1 than half and the others are way to big.

6 0

3 years ago

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Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental c

Elina [12.6K]

Answer:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

Step-by-step explanation:

Data given

$n_1 =8$ represent the sample size for group Boston

$n_2 =9$ represent the sample size for group Dallas

$\bar X_1 =47$ represent the sample mean for the group Boston

$\bar X_2 =44$ represent the sample mean for the group Dallas

$s_1=3$ represent the sample standard deviation for group Boston

$s_2=3$ represent the sample standard deviation for group Dallas

We can assume that we have independent samples from two normal distributions with equal variances and that is:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t student distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$\S^2_p =\frac{(8-1)(3)^2 +(9 -1)(3)^2}{8 +9 -2}=9$

And the deviation would be just the square root of the variance:

$S_p=3$

The statitsic would be:

$t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058$

The degrees of freedom are:

$df=8+9-2=15$

And the p value would be:

$p_v =P(t_{15}>2.058) =0.0287$

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are signiﬁcantly higher than those in Dallas

5 0

3 years ago