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3 years ago

6

Does anyone know this? if so would you be kind to help me out?

2 answers:

Neko [114]3 years ago

6 0

The answer is obtuse

Naya [18.7K]3 years ago

3 0

It is a right angle just right angle

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Carlos drank 2740 mL of water after football practice. How many liters did he drink

Bas_tet [7]

2740/1000= 2.74

So the answer is 2.74 Liters

4 0

3 years ago

Read 2 more answers

This my test question I'm 100 bad at math 2b2+16b+49÷2b+4

krek1111 [17]

2b^2 + 81/2 b + 4

basically 2b^2 + 16b + 49/2 b + 4
= (2b^2) + 16b+49/2b) + (4)
= 2b^2 + 81/2 b + 4

5 0

3 years ago

What's the smallest possible perimeter of a rectangle with an area of 35 cm? (36 small squares 1 cm by 1 cm)

vfiekz [6]

Answer:

12 miles per hour...........

6 0

3 years ago

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.

aleksandr82 [10.1K]

Answer:

The probability is 0.508 = 50.8%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.0525 g.

This means that $\mu = 0.8544, \sigma = 0.0525$

If 1 candy is randomly selected, find the probability that it weighs more than 0.8535 g.

This is 1 subtracted by the pvalue of Z when X = 0.8535. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.8535 - 0.8544}{0.0525}$

$Z = -0.02$

$Z = -0.02$ has a pvalue of 0.492

1 - 0.492 = 0.508

The probability is 0.508 = 50.8%.

8 0

3 years ago

The endpoints of a diameter of a circle are (6,2) and (−2,5) . What is the standard form of the equation of this circle?

Paul [167]

Answer:

The standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

Step-by-step explanation:

Given that the end points of a diameter of a circle are (6,2) and (-2,5);

Now to find the standard form of the equation of this circle:

The center is (h,k) of the circle is the midpoint of the given diameter

midpoint formula is $M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

Let $(x_{1},y_{1})$ and $(x_{2},y_{2})$ be the given points (6,2) and (-2,5) respectively.

$M=(\frac{6-2}{2},\frac{2+5}{2})$

$M=(\frac{4}{2},\frac{7}{2})$

$M=(2,\frac{7}{2})$

Therefore the center (h,k) is $(2,\frac{7}{2})$

now to find the radius:

The diameter is the distance between the given points (6,2) and (-2,5)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$=\sqrt{(-2-6)^2+(5-2)^2}$

$=\sqrt{(-8)^2+(3)^2}$

$=\sqrt{64+9}$

$=\sqrt{73}$

Therefore the radius is $\sqrt{73}$

i.e., $r=\sqrt{73}$

Therefore the standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Now substituting the center and radiuswe get

$(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2$

$(x-2)^2+(y-\frac{7}{2})^2=73$

Therefore the standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

5 0

3 years ago