All categories

3 years ago

(transition to college mathematics.)

Mathematics

1 answer:

prohojiy [21]3 years ago

5 0

Answer:

False

Step-by-step explanation:

Hope this helped

stay safe

Brainliest is appreciated i only need 1 more to level up please help :)))

You might be interested in

18,000 ÷ 60<br> = 18,000 ÷ 10<br> ÷ 6<br> =___÷ 6

ludmilkaskok [199]

Answer:

1800 ÷ 6

Step-by-step explanation:

18,000 ÷ 60 is the same as;

18,000 ÷ 10 ÷ 6 and is the same as;

1,800 ÷ 6

7 0

4 years ago

Select all of the questions for which x=3 is a solution

Brrunno [24]

Answer:

there are no questions

Step-by-step explanation:

5 0

3 years ago

henry is mixing concrete. to make concrete he needs to mix cement, sand and gravel in the ratio 5 : 4 : 1 by weight. henry needs

MaRussiya [10]

Answer:

cement by 5 Kg

Step-by-step explanation:

add the parts of the ratio, 5 + 4 + 1 = 10 parts

Divide the quantity of concrete by 10 to find the value of one part of the ratio.

110Kg ÷ 10 = 11Kg ← value of 1 part of ratio , thus

5 parts = 5 × 11Kg = 55Kg ← cement required

4 parts = 4 × 11Kg = 44Kg ← sand required

1 part = 11Kg ← gravel required

He requires 55Kg of concrete but only has 50Kg.

He requires 44Kg of sand and has 55Kg

He requires 11Kg of gravel and has 15Kg

Thus he is 5Kg short of cement.

7 0

3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago

What is the arc length of a 7cm diameter semi circle?

bija089 [108]

Answer:

11 cm

just find the ½ of the circumference of the full circle

3 0

3 years ago