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umka2103 [35]
2 years ago
15

What is 1 + 57 = ? help this is hard!

Mathematics
2 answers:
blagie [28]2 years ago
6 0

Answer:

58 LOL or is it 109? I don't know it's too hard for me to figure out but I think it's 58

Anvisha [2.4K]2 years ago
3 0

79 , this took some time but I figured it out

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Help me Please?
k0ka [10]

Answer:

f(9.5) = 558.02.

Step-by-step explanation:

Because it is an exponential function, it can be written: y = (a)(b)^x

Now substitute in the two known points:

72 = (a)(b)^7.5 and

2 = (a)(b)^4 : => since 2 = (a)(b)^2 divide the left side by 2 and the right by (a)(b)^2

72/2 =[(a)(b)^7.5]/[(a)(b)^4]: => the (a) cancels out and b becomes b^(7.5–4)

36 = b^3.5: => To isolate b take the 3.5 root of both sides

{3.5 root}(36) = {3.5 root}(b^3.5)

2.783927 = b

Now to solve for a

2 = (a)(b)^4 => 2 = (a)(2.783927)^4 => 2 = (a)(60.066375) divide both sides by 60.066375 0.0332965 = a

Equation: y = (0.0332965)(2.783927)^x

Now plug in 9.5 for x

(0.0332965)(2.783927)^9.5 = 558.0178525 = 558.02

7 0
3 years ago
A carnival charges $5 for admission plus $2 per ride.
almond37 [142]
1. y= 2r+5 2. $17 3. 5 rides
7 0
3 years ago
Can you please help me out
Ahat [919]

Answer:

yes I can. what is the question?

4 0
3 years ago
Please and thank you
ElenaW [278]
60 deg is 1/6 of 360 deg
So, 1/6 x pi5^2
= Approx. 13.1
A.

BRAINLIEST PLS!
5 0
4 years ago
Read 2 more answers
In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
pickupchik [31]

Answer:

The population standard deviation is not known.

90% Confidence interval by T₁₀-distribution: (38.3, 53.7).

Step-by-step explanation:

The "standard deviation" of $14 comes from a survey. In other words, the true population standard deviation is not known, and the $14 here is an estimate. Thus, find the confidence interval with the Student t-distribution. The sample size is 11. The degree of freedom is thus 11 - 1 = 10.

Start by finding 1/2 the width of this confidence interval. The confidence level of this interval is 90%. In other words, the area under the bell curve within this interval is 0.90. However, this curve is symmetric. As a result,

  • The area to the left of the lower end of the interval shall be 1/2 \cdot (1 - 0.90)= 0.05.
  • The area to the left of the upper end of the interval shall be 0.05 + 0.90 = 0.95.

Look up the t-score of the upper end on an inverse t-table. Focus on the entry with

  • a degree of freedom of 10, and
  • a cumulative probability of 0.95.

t \approx 1.812.

This value can also be found with technology.

The formula for 1/2 the width of a confidence interval where standard deviation is unknown (only an estimate) is:

\displaystyle t \cdot \frac{s_{n-1}}{\sqrt{n}},

where

  • t is the t-score at the upper end of the interval,
  • s_{n-1} is the unbiased estimate for the standard deviation, and
  • n is the sample size.

For this confidence interval:

  • t \approx 1.812,
  • s_{n-1} = 14, and
  • n = 11.

Hence the width of the 90% confidence interval is

\displaystyle 1.812 \times \frac{14}{\sqrt{10}} \approx 7.65.

The confidence interval is centered at the unbiased estimate of the population mean. The 90% confidence interval will be approximately:

(38.3, 53.7).

5 0
3 years ago
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