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Morgarella [4.7K]

3 years ago

11

Mrs.drew want to build a square sandbox with an area of 361 square feet what is the total length of wood mrs. Drew needs to make

the slide of the sandbox

1 answer:

Taya2010 [7]3 years ago

4 0

Answer:

180.5 ft

Step-by-step explanation:

the area for the sand box would be calculated as length × width. since the sandbox is square each side is the same length. the side will be half the number of the area.

361 ÷ 2 = 180.5 ft

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2/7 • 9 1/10<br> Please help

nordsb [41]

$\frac{2}{7} * 9 \frac{1}{10} = \frac{2}{7} * \frac{91}{10} = \frac{2~*~91}{7~*~10} = \frac{182}{70} = 2 \frac{42}{70} = 2 \frac{3}{5}$

3 0

3 years ago

Read 2 more answers

A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa

diamong [38]

Answer:

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}= \mu_D$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C,D$

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have $p=4$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

8 0

3 years ago

There are currently 3 people signed up for the debate club. The school will not recognize the club officially until it has more

svetlana [45]

Answer:

Number line with open circle on 2 and shading to the right

Step-by-step explanation:

this is showing that it has to be greater than two and is going up on the number line

7 0

2 years ago

1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help

Gnesinka [82]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Simplify :- 1 + - w² + 9w.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\large \sf1 + - w ^ { 2 } + 9 w$

Quadratic polynomial can be factored using the transformation $\sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ , where $\sf x_{1} and x_{2}$ are the solutions of the quadratic equation $\sf \: ax^{2}+bx+c=0$ .

$\large \sf-w^{2}+9w+1=0$

All equations of the form $\sf\:ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}$ . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)} \\$

Square 9.

$\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)} \\$

Multiply -4 times -1.

$\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)} \\$

Add 81 to 4.

$\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)} \\$

Multiply 2 times -1.

$\large \sf \: w=\frac{-9±\sqrt{85}}{-2} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is plus. Add -9 to $\sf\sqrt{85}$ .

$\large \sf \: w=\frac{\sqrt{85}-9}{-2} \\$

Divide -9+ $\sf\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is minus. Subtract $\sf\sqrt{85}$ from -9.

$\large \sf \: w=\frac{-\sqrt{85}-9}{-2} \\$

Divide $\sf-9-\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}} \\$

Factor the original expression using $\sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ . Substitute $\sf\frac{9-\sqrt{85}}{2}$ for $\sf\:x_{1}$ and $\sf\frac{9+\sqrt{85}}{2}$ for $\sf\:x_{2}$ .

$\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}$

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

$\large \sf \: 1 + - w {}^{2} + 9w \\ =\large \boxed{\bf \: 1 - {w}^{2} + 9w}$

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0

2 years ago

How many milliliters of water are in each 7 beakers

mr Goodwill [35]

Can you ask the full question please?

7 0

3 years ago