All categories

2 years ago

Can someone help solve? will give brainliest

Mathematics

1 answer:

emmainna [20.7K]2 years ago

8 0

Answer:

Step-by-step explanation:

Your answer

About all the sides are equal so we can say (10x + 44) side's triagle is twice than another triangle

So, 10x + 44 = 2( 8x - 23)

..... 10x +44 = 16x -46

...... X = 15

Mark it as Brainlist answer. Follow me.

You might be interested in

Mr. George has a bank balance of -42 dollars at the start of the month. After he deposits 6 dollars, the new balance is (with fu

OLga [1]

Answer:

$-36

Step-by-step explanation:

-42+6

-36

5 0

2 years ago

I WILL GIVE A BRANLIEST!<br>What is the graph of the functions?<br> y=2*2^x<br> y=-2*5^x

tekilochka [14]

For this case what you should know is that both functions are of the potential type.
We have then that
y = 2 * 2 ^ x
This function grows exponentially upwards.
y = -2 * 5 ^ x
This function grows exponentially downwards.
Answer
See attached graphics.

4 0

2 years ago

What happened to rectangle A to become rectangle B in the diagram below?

Bumek [7]

Answer: Dilated and then translated

Step-by-step explanation:

8 0

1 year ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$

8 0

2 years ago

What is the lowest common denominator of the following rational expressions?

Damm [24]

Answer:

I think its A

Step-by-step explanation:

3 0

2 years ago