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Nadya [2.5K]
3 years ago
14

7x + 17 100° Solve for X

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

7x =100-17

7x = 83

X= 11/6/7

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LOTS OF POINTS!!!<br> SOLVE ANd EXPLAIN PROBLEM 8
alexandr1967 [171]

Answer:

r≈2.26in

Step-by-step explanation:

3 0
3 years ago
Factor -1/2out of -1/2x+6
Volgvan
-\dfrac{1}{2}x+6=-\dfrac{1}{2}\left(\dfrac{-\frac{1}{2}x}{-\frac{1}{2}}+\dfrac{6}{-\frac{1}{2}}\right)=-\dfrac{1}{2}\left(x-\dfrac{6\cdot2}{1}\right)=\boxed{-\dfrac{1}{2}\left(x-12\right)}
3 0
3 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
3 years ago
FONTS
Wewaii [24]

Answer:

x=5

Step-by-step explanation:

9x^2 - 2x + 25 = 8x^2 + 8x

9x^2-8x^2-2x-8x+25=0

x^2-6x+25=0 factorize

(x-5)(x-5)=0

x-5=0 then x=5

8 0
3 years ago
The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
2 years ago
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