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Dimas [21]
3 years ago
9

I need help !!!!!!!!!!!!!

Mathematics
1 answer:
hoa [83]3 years ago
8 0

Answer:

see below

Step-by-step explanation:

7 x 13    (split 13 into 10 + 3)

= 7 x ( 10 + 3)     (expand parentheses by distributive property)

= (7 x 10) + (7 x 3)

= 70 + 21

= 91

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Khalil with me six times as many baskets at John during basketball practice
ycow [4]

Answer:

Step-by-step explanation:

how many baskets does john have?

8 0
3 years ago
For the class of 2013's prom Norman's dress shop sold cheaper dresses for $90 each and more expensive dresses for $140 each. The
Tema [17]

Answer:

Number of cheaper dresses sold  is 35

Number of expensive  dresses sold  is 15

Step-by-step explanation:

Given:

Cost of cheaper dresses = $90

Cost of  expensive dresses = $140

Total cost of  the dresses = $5250

To Find:

Number of cheaper dress = ?

Number of expensive  dress = ?

Solution:

Let

The number of  cheaper dresses be x

The number of  expensive dresses be y

(Number of cheaper dresses X cost of cheap dress) +  (Number of Expensive dresses X cost of  expensive dress)  =  $5250

x \times90 +y \times 140 = 5250=  $5250

It is given that the 20 more of the cheaper dresses than the expensive dresses is sold

So,

number of cheaper dress  =  20  +  number of expensive dress

x = 20 + y---------------------------------------(1)

(20+y) \times90 +y \times 140 = 5250 = 5250

(20 \times 90 +y\times 90) +y \times 140= 5250

1800 + 90y+ 140y = 5250

1800 + 230y = 5250

230y =5250 -1800

230y = 3450

y = \frac{3450}{230}

y = 15

Substituting y in (1)

x = 20 +15

x= 35

5 0
4 years ago
How do you solve 5-x=6?
Fudgin [204]
-x+5-5=6-5, -x=1, x=-1
3 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
Can somebody help plz​
Evgen [1.6K]

Answer:

(3,2)

Step-by-step explanation:

the other three ordered pairs do not lie in the shaded region

6 0
3 years ago
Read 2 more answers
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